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Measuring thermal quantities by the method of mixtures - class-XI

Description: measuring thermal quantities by the method of mixtures
Number of Questions: 74
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Tags: physics thermal physics heat calorimetry heat and thermodynamics heat and temperature thermal properties thermal properties of matter properties of bulk matter measurement and effects of heat
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How much heat is required to raise the temperature of 150 g of iron from $20 ^oC$ to $25 ^oC$? (Specific heat of iron $480 J kg^{-1} {\;}^oC^{-1})$

  1. 350 J

  2. 345 J

  3. 360 J

  4. 330 J


Correct Option: C
Explanation:

Given $m=150$
$g=\frac {150}{1000}=0.15 kg$
Specific heat of iron
$C=480 J kg^{-1} {\;}^oC^{-1})$
$\Delta =(25-20)^oC=5^oC$
$Q=m\times C\times \Delta T$
$=0.15 kg\times 480 J kg^{-1} {\;}^oC^{-1}\times 5^oC$
$=360 J$.

Find the heat lost by a copper cube of mass 400 g when it cool from $100^oC$ to $30^oC$. (Specific of heat of copper $=390 J kg^{-1} {\;}^oC^{-1})$.

  1. 50000 J

  2. 10000 J

  3. 10920 J

  4. 10900 J


Correct Option: C
Explanation:

$m=400 g=0.4 kg$
$C=390 J kg^{-1} {\;}^oC^{-1}$
$T _1=100 ^oC, T _2=30^oC$
$\Delta T=70^oC$
$\therefore$ Heat lost $=mC(T _1-T _2)$
$=0.4\times 390\times 70 J=10,920 J$

A body having 1680 J of energy is supplied to 1000 g of water. If the entire amount of energy is converted into heat, the rise in temperature of water (sp. heat of water $=4200 J kg^{-1} {\;}^oC^{-1})$

  1. $0.4 ^oC$

  2. $40 ^oC$

  3. $4 ^oC$

  4. $44 ^oC$


Correct Option: A
Explanation:

$Q=1680 J$
$m=1000 g$ or 1 kg


$\Delta T=\dfrac {Q}{mc}$

$=\dfrac {1680}{1\times 4200}=0.4^oC.$

What is the process in which heat energy of both hot and cold body equalizes?

  1. Calorimetry

  2. Fermentation

  3. Latent heat

  4. Hidden heat


Correct Option: A
Explanation:

Principle of calorimetry states that heat lost by a hotter body = heat gained by a colder body, therefore calorimetry is a process in which heat energy of both hot and cold body equalizes.

How much heat is required to raise the temperature of 100 g of water of $5 ^oC$ to $95 ^oC$?

  1. 900 kcal

  2. 90 kcal

  3. 10 kcal

  4. 9 kcal


Correct Option: D
Explanation:

$m=100 g, C=1 cal g^{-1} {\;}^oC^{-1}$
$\Delta T=(95-5)^oC=90^oC$
$Q=mC\Delta T$
$=100 g\times 1 cal g^{-1} {\;}^oC^{-1}\times 90^oC$
$=100\times 1\times 90 cal$
$=9000 cal=9 k cal$.

According to principle of calorimetry, heat absorbed by cold bodies is equal to heat released by hot bodies.

  1. True

  2. False


Correct Option: A
Explanation:
The statement is true, according to principle of calorimetry, heat absorbed by cold bodies is equal to heat released by hot bodies.
It can also be understood by conservation of energy concept.

5 kg of water at $80^oC$ is taken in a bucket of negligible heat capacity, 15 kg of water at $20^oC$ is added to it. What is the temperature of the mixture?

  1. $45^oC$

  2. $65^oC$

  3. $85^oC$

  4. $35^oC$


Correct Option: D
Explanation:

Hot water, $m _h=5 kg, T _h=80^oC$
Cold water, $m _c=15 kg, T _c=20^oC$
$T=?$
If the temperature of mixture is T
Heat lost by hot water
$5\times C\times (80-T)$
Heat gained by cold water
$15\times C\times (T-20)$
According to the principle of calorimetry, Heat lost $=$ Heat gained
$5\times C\times (80-T)=15\times C\times (T-20)$
$\therefore 80-T=3(T-20)$

$T=35^oC$.

5 g of water at $30^oC$ and 5 g of ice at $-20^oC$ are mixed together in a calorimeter. What is the final temperature of the mixture. Given specific heat of ice $=0.5 cal g^{-1} (^oC)^{-1}$ and latent heat of fusion of ice $=80 cal g^{-1}$.

  1. $0^oC$

  2. $1^oC$

  3. $10^oC$

  4. $20^oC$


Correct Option: A
Explanation:

Heat lost by 5 g of water at $30^oC$ to water at $0^oC$.
$Q _L=5\times 1\times 30=150 cal$
Heat required by 5 g of ice at $-20^oC$
$(Q _1)=5\times 0.5\times (2.0)=50 cal$
Heat required by 5 g of ice at $0^oC$  into water at $0^oC$.
$(Q _2)=5\times 80=400 cal$
$Q _L < (Q _1+Q _2)$
$\therefore$ The final temperature of the mixtuure is $0^oC$.

In a process 10 g of ice at $-5^oC$ is converted into the steam at $100^oC$. If specific heat of ice is $0.5 \ cal g^{-1} {\;}^oC^{-1}$, then the amount of heat required to convert 10 g of ice from $-5^oC$ to $0^oC$ is :

  1. 15 cal

  2. 25 cal

  3. 50 cal

  4. 100 cal


Correct Option: B
Explanation:

$C=0.5 cal g^{-1} {\;}^oC^{-1}$


Heat required to rise the temperature of ice at $-5^oC$ to $0^oC$ is

$Q=mC\Delta T$

$=(10g)(0.5 \ cal g^{-1} {\;}^oC^{-1})$$(0^oC)-(-5^oC)$

$=25 cal$

400 g of ice at 253 K is mixed with 0.05 kg of steam at $100^oC$. Latent heat of vaporisation of steam $=540 cal g^{-1}$. Latent heat of fusion of ice $=80 cal g^{-1}$. Specific heat of ice $=0.5 cal g^{-1} {\;}^oC^{-1}$. Find the resultant temperature of the mixture.

  1. 253 K

  2. 260 K

  3. 273 K

  4. 290 K


Correct Option: C
Explanation:

Heat lost by 0.05 kg of steam at $100^oC$ to water at $0^oC$
$Q _L=(50\times 540)+(50\times 1\times 100)$
$=27000+5000=32000 cal$
Heat required by 400 g of ice at 253 K $(-20^oC)$ to convert into water at 273 K $10^oC) Q _1=(400\times 0.5\times 20)+(400\times 80)$
$=(4000+32000)=36000 cal$
$\therefore Q _L < Q _1$, only part of ice melts and final temperature remains $0^oC$ or 273 K.

A piece of copper weighing 500 g is heated to $100^oC$ and dropped into 200g of water at $25^oC$. Find the temperature of the mixture. The specific heat of Cu is $0.42 J g^{-1} {\;}^oC^{-1}$.

  1. $30^oC$

  2. $40^oC$

  3. $50^oC$

  4. $60^oC$


Correct Option: B
Explanation:

Given, mass of copper, $m _1=500 g$
Mass of water, $m _2=200 g$
Initial temperature of copper, $t _1=100^oC$
Initial temperature of water, $t _2=25^oC$
Sp. heat of copper, $C _1=0.42 J g^{-1} {\;}^oC^{-1}$
Sp. heat of water, $C _2=4.2 J g^{-1} {\;}^oC^{-1}$
Final temperature of the mixture $=t^oC$
Then,
Heat lost by the copper piece
$=m _1C _1(t _1-t)$
Heat gained by water $=m _2C _2(t-t _2)$
We know, Heat lost $=$ Heat gained
$\Rightarrow m _1C _1(t _1-t)=m _2C _2(t-t _2)$
$\Rightarrow 500\times 0.42\times (100-t)$
$=200\times 4.2\times (t-25)$
$\Rightarrow (100-t)=\frac {200\times 4.2}{500\times 0.42}\times (t-25)$
$=4(t-25)$
This given, $5t=200$
$\Rightarrow t=\frac {200}{5}^oC=40^oC$
Thus, the final temperature of the mixture is $40^oC$.

$10\ kg$ of hot water in a bucket at $70^oC$ is cooled for taking a bath adding to it $20\ kg$ water at $20^oC$. What is the temperature of the mixture? (Neglect the thermal capacity of the bucket)

  1. $30.67^oC$

  2. $36.67^oC$

  3. $60.67^oC$

  4. $46.67^oC$


Correct Option: B
Explanation:

m (hot water) $=10 kg$,
T (hot water) $=70^oC$
m (cold water) $=20 kg$
T (cold water) $=20^oC, T (final)=?$
Using the formula $Q=mC\Delta t$
We get heat lost by hot water
$=10\times C\times (70-T _f)$
Where $T _f$ is the final temperature
Heat gained by cold water
$=20\times C\times (T _f-20)$
Using the principle of calorimetry
Heat lost $=$ Heat gained
We get $10\times C\times (70-T _f)$
$=20\times C\times (T _f-20)$
$\therefore 700-10T _f=20T _f-400$
or $30T _f=1100 \therefore T _f=36.67^oC$.

What is the final temperature of the mixture of 300 g of water at $25^oC$ added to 100 of ice at $0^oC$.

  1. $0^oC$

  2. $1^oC$

  3. $2^oC$

  4. $3^oC$


Correct Option: A
Explanation:

Heat lost by a hot body $=$ Heat gained by a

$\therefore 300 (25-\theta)=100\times 80+100\times 0.5\theta$

$\therefore \theta=-\frac {5}{3.5}$

Since $\theta$ is negative

Heat lost is utilised to melt only same part of ice. Hence equilibrium temperature is $0^oC$.

500 g of water at $100^oC$ is mixed with 300 g at $30^oC$. Find the temperature of the mixture. Specific heat of water $=4.2 J g^{-1} {\;}^oC^{-1}$.

  1. $73.8^oC$

  2. $53.8^oC$

  3. $40^oC$

  4. $60^oC$


Correct Option: A
Explanation:

Mass of hot water, $m _1=500 g$
Mass of cold water, $m _2=300 g$
Temp. of hot water, $t _1=100^oC$
Temp. of cold water, $t _2=30^oC$
Sp. heat of water, $C=4.2 J g^{-1} {\;}^oC^{-1}$
Let temp. of mixture be $t^oC$. Then, Heat gained by cold water
$=m _2\times C\times (t-t _2)$
According to the principle of calorimetry, Heat lost $=$ Heat gained
$500\times 4.2\times (100-t)$
$=300\times 4.2\times (t-30)$
$\Rightarrow 5(100-t)=3(t-30)$
$\Rightarrow -3t-5t=-90-500$
$\Rightarrow -8t=-590$
$\Rightarrow t=\frac {590}{8}=73.8^oC$
So, the final temperature of the mixture is $73.8^oC$.

A calorimeter contains $70.2 \,g$ of water at $15.3^o C$. If $143.7 \,g$ of water at $36.5^o C$ in mixed it with the common temperature is $28.7^o C$. The water equivalent of the calorimeter is:

  1. $15.6 \,g$

  2. $9.4 \,g$

  3. $6.3 \,g$

  4. $13.4 \,g$


Correct Option: D
Explanation:

Assume water equilient is $= ngm$
So, $(70.2) \times 1 \times (28.7 - 15.3) + n (28.7 - 15.3) = 143.7 \times 1 (36.5 - 28.7)$
$n (13.4) = 180.18$
$n = 13.4 \,g$

$5$g of copper was heated from $20^{\circ}$ to $80^{\circ}$. How much energy was used to heat Cu? (Specific heat capacity of Cu is $0.092 cal/g ^{\circ}C$).

  1. $27.6$ cal

  2. $50$ cal

  3. $35$ cal

  4. $25.7$ cal


Correct Option: A
Explanation:
Given,

Mass, $m=5\,g$

Specific heat capacity, $C=0.092\,cal/g \,^0C$

Change in temperature, $\Delta T=80\,^0C -20^0C=600^0C$

Heat required, $Q=?$

We have the equation,

$Q=m\times C\times \Delta T$

Then,

$Q=5\times 0.092\times 60=27.6\,cal$

A calorimeter constains 10 g of water at ${ 20 }^{ \circ  }$ C. The temperature falls to ${ 15 }^{ \circ  }$ C in 10 min. When calorimeter contains 20 g of water at ${ 20 }^{ \circ  }$ C, it takes 15 min for the temperature to become ${ 15 }^{ \circ  }$ C. The water equivalent of the calorimeter is

  1. 5 g

  2. 10 g

  3. 25 g

  4. 50 g


Correct Option: C

What is the principle of the method of a mixture? Name the law on which this principle is based.

  1.  Newtons law of cooling 

  2. none

  3. principle of calorimetry 

  4. principle of heat transfer


Correct Option: C
Explanation:

A principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
i.e. heat loss = heat gained
It is based on Newton's Law of Cooling, which states that when a liquid is heated of higher temperature and placed to cool. Then the rate of heat lost by a temperature of the liquid is directly proportional to the difference in temperature of the surrounding.

In a calorimeter of water equivalent  $20 { g },$  water of mass  $1.1 { kg }$  is taken at  $288{ K }$  temperature. If steam at temperature  $373 { K }$  is passed through it and temperature of water increases by  $6.5 ^ { \circ } { C }$  then the mass of steam condensed is

  1. $17.5{ g }$

  2. $11.7{ g }$

  3. $15.7{ g }$

  4. $18.2{ g }$


Correct Option: B

2000 J of energy is needed to heat 1 kg of paraffin through $1^{\circ}C$. So How much energy is needed to heat 10 kg of paraffin through $2^{\circ}C$ ?

  1. 4000 J

  2. 10,000 J

  3. 20,000 J

  4. 40,000 J


Correct Option: D
Explanation:

$W= mc\theta$
$2000= (1000)c(1)$
$c= 2$ $J/g^oC$

we get value of c
Hence for $10 kg$ through $2^oC$,
$W= (10000)(2)(2)= 40000 J$

When in thermal contact, the quantity of heat lost by the hotter body is ...... the amount of heat gained by the colder body. (neglect loss of heat due to convection & radiation)

  1. Equal to

  2. Greater than

  3. Less than

  4. Cannot say


Correct Option: A
Explanation:
When in thermal contact, the quantity of heat lost by the hotter body is equal to the amount of heat gained by the colder body. (neglect loss of heat due to convection & radiation)
This is the principle of calorimetry.

Bunty mixed 440 gm of ice at $0^{\circ}C$ with 540 gm of water at $80^{\circ} C$ in a bowl. Then what would remain after sometime in the bowl?

  1. only ice

  2. only water

  3. ice and water in same amount

  4. ice and water will vapourise


Correct Option: B
Explanation:

Energy for ice= mL=440(336)= 147840 J
Energy in water= mc$\theta$= 540(80)(4.2)= 181440J

Since, water has more energy, the ice will completely melt while the temperature of water will decrease.
so only ice remains.

The quantity of heat required to raise the temperature of 2000 g of water from 10$^o$C to 50$^o$C is

  1. 80 cal

  2. 80,000 cal

  3. 8000 cal

  4. none of these


Correct Option: B
Explanation:

Heat required $=mS\triangle T \ =2000\times 1\times (50-10) \ =80000cal$

The quantity/ quantities that does/do not have mass in its/their dimensions is/are

  1. Specific heat

  2. Latent heat

  3. Electrical potential difference

  4. Electrical resistance


Correct Option: C,D
Explanation:

Specific heat- $J/kg^oC$
Latent Heat- J/kg
Electrical potential difference- J/coulumb
Electrical resistance- ohm
 Hence, option C and D are correct

When 60 calories of heat are supplied to 15 g of water, the rise in temperature is

  1. $75^{\circ}C$

  2. $90^{\circ}C$

  3. $4^{\circ}C $

  4. $0.25^{\circ}C$


Correct Option: C
Explanation:

$Heat = m S \Delta T$
$\Rightarrow 60=15\times 1\times\triangle T \ \Rightarrow \triangle T=4^oC$

Calorimeters are generally made of

  1. copper

  2. brass

  3. aluminium

  4. zinc


Correct Option: A
Explanation:
Calorimeters is made of copper because:
1) Copper has low specific heat capacity and thus it reaches equilibrium temperature quickly by absorbing small amount of heat.

2) Cu is highly malleable and hence very thin box of small mass can be made, thus decreasing heat capacity of box.

3) Cu is good conductor of heat.

On which law does the study of calorimetry based?

  1. Joule's law

  2. Law of conservation of energy

  3. Law of Kinetic energy

  4. None


Correct Option: B
Explanation:

Principle of calorimetry states that heat lost by a hotter body = heat gained by a colder body, therefore in calorimetry, total heat energy of the system remains constant, which is the law of conservation of energy . 

When 60 calories of heat are supplied to 15 g of water, the rise in temperature is

  1. $75^\circ C$

  2. $900^\circ C$

  3. $4^\circ C$

  4. $0.25^\circ C$


Correct Option: C
Explanation:

We know that, $heat=mass\times specific \;heat\times change\;in\;temperature \ \Rightarrow 60=15\times 1\times \triangle T \ \Rightarrow \triangle T=4^oC$

A thermos bottle containing coffee is vigorously shaken. If the coffee is considered as a system, then the temperature of the coffee will 

  1. increase slightly

  2. fall

  3. remain the same

  4. never be determined


Correct Option: A
Explanation:
A thermoflask is closed system, so no heat energy added to system i.e. $\triangle \theta=0$
Work is done on the system by shaking it against viscous force. Hence $\triangle w=-ve$
Internal energy, $\triangle U=\triangle \theta - \triangle w=0-(-\triangle w) \\ \Rightarrow \triangle U=+\triangle w$
$\because \triangle U\; is\; +ve$ So there is increase in temperature due to rise in internal enrgy.

The branch of physics that deals with the measurement of heat energy is known as

  1. Fermentation

  2. Latent heat

  3. Calorimetry

  4. Hidden heat


Correct Option: C
Explanation:

Calorimetry is the branch of physics which deals with the measurement of heat energy. Calorimetry is one of the methods for the determination of specific heats or latent heats of the substances.

A thermometer is used to measure 

  1. heat

  2. thermal capacity

  3. water equivalent

  4. temperature


Correct Option: D
Explanation:
A thermometer is used to measure temperature.
Thermo(heat)             meter(measuring device)
It measures temperature in either celcius or farenheit scale.

Which of the following properties must be known in order to calculate the amount of heat needed to melt 1.0kg of ice at $0^oC$? 
I. The specific heat of water 
II. The latent heat of fusion for water 
III. The density of water.

  1. I only

  2. I and II only

  3. I, II, and III

  4. II only

  5. I and III only


Correct Option: D
Explanation:

The latent heat is the heat required to change the state of unit mass of substance ,  therefore heat required to change the mass $m$ of substance  is given by ,

            $Q=mL$ ,  where $m=$ mass of substance , $L=$ latent heat
 here we have $m=1.0kg$ but we don't have value of $L$ (latent heat of fusion for water) so it is required .
    Density and specific heat of water are not required here , as it is clear from formula mentioned above .

Heat is added to a block of ice of mass $m$ until the entire block melts into liquid water. Identify by which of the following method this can be explained ?

  1. First law of thermodynamics (conservation of energy)

  2. Second law of thermodynamics (law of entropy)

  3. Ideal gas law

  4. Heat of fusion and heat of vaporization equation

  5. Heat engine efficiency


Correct Option: D
Explanation:

When heat is added to ice at $0^{\circ}C$, the temperature of the ice does not change. However the heat goes into the latent heat of fusion of ice, This is a kind of potential energy that water owns in form of latent heat.

Hence correct answer is option D.

State True or False.


According to principle of calorimetry heat absorbed by cold bodies is equal to heat released by hot bodies.

  1. True

  2. False


Correct Option: A
Explanation:
True
According to principle of calorimetry; heat absorbed by cold bodies is equal to heat released by hot bodies. Heat flows from a body at higher temperature to body at lower temperature. Heat will transfer till bodies come in thermal equilibrium that is, they reach at the same temperature. And heat released is equal to absorbed if no heat is dissipated to surrounding.

400 g of vegetable oil of specific heat capacity 1.98 J ${ g }^{ -1 }$ $^{ \circ  }{ { C }^{ -1 } }$) is cooled from ${ 100 }^{ \circ  }C$. Find the final temperature, if the heat energy given out by is 47376 J.

  1. ${ 30.2 }^{ \circ }C$

  2. ${ 40.2 }^{ \circ }C$

  3. ${ 50.2 }^{ \circ }C$

  4. ${ 43.2 }^{ \circ }C$


Correct Option: B
Explanation:

Given ,  $m=400g ,  \theta _{1}=100^{0}C , \theta _{2}=?$ , specific heat of  vegetable oil $c=1.98J/g-^{o}C , Q=47376J$

Now ,  by the definition of specific heat c ,
                    $Q=mc\Delta \theta=mc(\theta _{1}-\theta _{2})$
or                 $47376=400\times1.98(100-\theta _{2})$
or                 $(100-\theta _{2})=47376/(400\times1.98)=59.8$
or                 $\theta _{2}=100-59.8=40.2^{o}C$

How much heat is required to raise the temperature of $150 g$ of iron from ${ 20 }^{ \circ  }C$ to ${ 25 }^{ \circ  }C$?

  1. $350 J$

  2. $345 J$

  3. $360 J$

  4. $330 J$


Correct Option: B
Explanation:

Given ,  $m=150g ,  \theta _{1}=20^{0}C , \theta _{2}=25^{0}C$

We have , specific heat of iron $c=0.46J/g-^{o}C$
Now , heat required to raise the temperature of iron is given by the definition of specific heat c ,
                    $Q=mc\Delta \theta=mc(\theta _{2}-\theta _{1})$
or                 $Q=150\times0.46\times(25-20)=345J$

How much heat is required to raise the temperature of $100 g$ of water from ${ 5 }^{ \circ  }C$ to ${ 95 }^{ \circ  }C$?

  1. $900 kcal$

  2. $90 kcal$

  3. $10 kcal$

  4. $9 kcal$


Correct Option: D
Explanation:

Given ,  $m=100g ,  \theta _{1}=5^{0}C , \theta _{2}=95^{0}C$

We have , specific heat of water $c=1cal/g-^{o}C$
Now , heat required to raise the temperature of water is given by ,
                    $Q=mc\Delta \theta=mc(\theta _{2}-\theta _{1})$
or                 $Q=100\times1\times(95-5)=9000cal=9kcal$

2000 cal of heat is supplied to 200 g of water. Find the rise in temperature. (Specific heat of water = 1 cal ${ { g }^{ -1 } }^{ \circ  }{ C }^{ -1 }$)

  1. ${ 10 }^{ \circ }C$

  2. ${ 20 }^{ \circ }C$

  3. ${ 30 }^{ \circ }C$

  4. ${ 40 }^{ \circ }C$


Correct Option: A
Explanation:

Given ,  $m=200g , Q=2000cal ,  \Delta\theta=? , $ , specific heat of water $c=1cal/g-^{o}C$

Now , heat required to raise the temperature of water is given by the definition of specific heat ,
                    $Q=mc\Delta \theta$
or                 $\Delta \theta=Q/(mc)=2000/(200\times1)=10^{o}C$

What will be the amount of heat required to convert $50 g$ of ice at ${ 0 }^{ \circ  }C$ to water at ${ 0 }^{ \circ  }C$?

  1. $400 cal$

  2. $4000 cal$

  3. $3000 cal$

  4. $300 cal$


Correct Option: B
Explanation:

Amount of heat required to convert unit mass of ice into water is called latent heat ($L$) of fusion of ice  i.e.

                       $Q=mL$ ,
   given ,          $m=50g$ ,

   we have ,     $L=80cal/g$

Hence ,           $Q=50\times80=4000cal$ 

Calculate the quantity of heat required to convert 1.5 kg of ice at ${ 100 }^{ \circ  }C$ to water at ${ 15 }^{ \circ  }C$. (${ L } _{ ice }\quad =\quad 3.34\quad \times \quad { 10 }^{ 5 }\quad J{ \quad kg }^{ -1 }$, ${ C } _{ water }\quad =\quad 4180\quad J{ \quad kg }^{ -1 }\quad ^{ \circ  }{ { C }^{ -1 } }$)

  1. $5.85\quad \times \quad { 10 }^{ 5 }\quad J$

  2. $5.95\quad \times \quad { 10 }^{ 5 }\quad J$

  3. $3.95\quad \times \quad { 10 }^{ 5 }\quad J$

  4. $4.95\quad \times \quad { 10 }^{ 5 }\quad J$


Correct Option: B
Explanation:

We have ,  $L _{ice}=3.34\times10^{5}J/kg , m=1.5kg$

From the definition of latent heat , heat required to convert ice into water at constant temperature $0^{o}C$ ,
             $Q _{1}=mL _{ice}$
or          $Q _{1}=1.5\times3.34\times10^{5}=5.01\times10^{5}J$
Now , heat required to heat up the water at $0^{o}C$ to$15^{o}C$ ,
              $Q _{2}=mc(15-0)=1.5\times4180\times15=0.94\times10^{5}J$  , where $c=4180J/kg-^{o}C$
 Total heat required ,
             $Q=Q _{1}+Q _{2}$ 

or          $Q=5.01\times10^{5}+0.94\times10^{5}=5.95\times10^{5}J$ 

One calorie is defined as the heat required to raise the temperature of $1$ gm of water by $1^o$C in a certain interval of temperature and at certain pressure. The temperature interval and pressure is?

  1. $13.5^o$ C to $14.5^o$ C & $76$ mm of Hg

  2. $6.5^o$ C to $7.5^o$ C & $76$ mm of Hg

  3. $14.5^o$ C to $15.5^o$ C & $760$ mm of Hg

  4. $98.5^o$ C to $99.5^o$ C & $760$ mm of Hg


Correct Option: C
Explanation:

One calories is defined as the amount of heat required to raise the temp of $1\ gm$ of water from $14.5^oC$ to $15.5^oC$

in $760\ mm$ of $Hg$.

If there are no heat losses to the surroundings, the quantity of heat gained by the cold body is equal to the quantity of heat lost by the hot body.

  1. True

  2. False


Correct Option: A
Explanation:
Yes, its true that if there are no heat losses to the surroundings, the quantity of heat gained by the cold body is equal to the quantity of heat lost by the hot body.
If there is no heat loss, as heat is a form of energy; by conservation of energy i.e. energy can neither be created nor destroyed but can be converted from one form to other or can be transferred from one body to other.

A copper ball of mass $100gm$ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100gm$, filled with $170gm$ of water at room temperature. Subsequently the temperature of the system is found to b4 ${75}^{o}$. $T$ is given by then (Given: room temperature $={30}^{o}C$, specific heat of copper $=0.1cal/gm _{  }^{ o }{ C }\quad $)

  1. ${ 825 }^{ o }C$

  2. ${ 800 }^{ o }C$

  3. ${ 885 }^{ o }C$

  4. ${ 1250 }^{ o }C$


Correct Option: C
Explanation:
Final temperature of celomiter and its constant is given as
$To=75^o C$
$\Rightarrow \ 100\times 0.1\times (75-T)+100\times 0.1(75-30)+1.70 \times 1\times (75.32)$
$\therefore \ T=885^oC$

400 g of vegetable oil of specific heat capacity $1.98 J g^{-1} {\;}^oC^{-1}$ is cooled from $100^oC$. Find the final temperature, if the heat energy given out by oil is 47376 J.

  1. $30.2^oC$

  2. $40.2^oC$

  3. $50.2^oC$

  4. $43.2^oC$


Correct Option: B
Explanation:

$m=400 g, C=1.98 Jg^{-1} {\;}^oC^{-1}$
$T _1=100^oC, T _2=?$
Fall in temperature
$\Delta T=(100-x)$
Heat energy given out by oil
$=47376 J$
According to formula $Q=m.C.\Delta T$
$\Rightarrow 47376=400\times 1.98 (100-x)$
$\Rightarrow 100-x=\frac {47376}{400\times 1.98}=59.8$
$\Rightarrow x=100-59.8=40.2^oC$
$\therefore$ Final temperature of oil $=40.2^oC$.

2000 cal of heat is supplied to 200 g of water. Find the rise in temperature. (Specific heat of water $=1 cal g^{-1} {\;}^oC^{-1})$

  1. $10 ^oC$

  2. $20 ^oC$

  3. $30 ^oC$

  4. $40 ^oC$


Correct Option: A
Explanation:

Quantity of heat supplied
$Q=2000 cal$
Mass of water, $m=200 g$
Specific heat of water
$C=1 cal g^{-1} {\;}^oC^{-1}$
Rise in temperature $=?$
From relation, $Q=m C\Delta T$
$\Rightarrow \Delta T=\frac {Q}{m.C}$
$=\frac {2000 cal}{200 g\times 1cal g^{-1} {\;}^oC^{-1}}=10^oC$
So, the temperature of water rises by $10^oC$.

500 g of hot water at $60^oC$ is kept in the open till its temperature falls to $40^oC$. Calculate the heat energy lost to the surroundings by the water. (Specific heat of water $=4200 J kg^{-1} {\;}^oC^{-1})$

  1. 2400 J

  2. 5000 J

  3. 40000 J

  4. 42000 J


Correct Option: D
Explanation:

$m=500, g=0.5 kg$
$\Delta T=60-40=20^oC$
$C=4200 J kg^{-1} {\;}^oC^{-1} Q=?$
Using the formula, $Q=m C \Delta T$
$=0.5\times 4200\times 20=42000 J$
$\therefore \text {Heat lost}=42,000 J$

How much amount of heat is required to raise the temperature of 100 g of water from $30 ^oC$ to $100 ^oC$? The specific heat of water $=4.2 J g^{-1} {\;}^oC^{-1}$.

  1. 25.5 kJ

  2. 29.4 kJ

  3. 30 kJ

  4. 40 kJ


Correct Option: B
Explanation:

Mass of water, $m=100 g$
Rise in temperature $(\Delta T)$
$=(100^oC-30^oC)=70 ^oC$
Specific heat of water
$C=4.2 J g^{-1} {\;}^oC^{-1}$
Then $Q=m . C. \Delta T$
$=100 g\times 4.2 J g^{-1} {\;}^oC^{-1}\times 70 ^oC$
$=29400 J=29.4 kJ$

What quantity of heat would be given out by 200 gm of copper in cooling from $80^oC$ to $20^oC$ (Specific heat of copper $=0.09 cal g^{-1} {\;}^oC^{-1})$?

  1. 1080 cal

  2. 1000 cal

  3. 1500 cal

  4. 1100 cal


Correct Option: A
Explanation:

$m=200 g, C _{copper}=0.09 cal g^{-1} {\;}^oC^{-1}$
$\Delta T=T _1-T _2=80-20=60^oC$
$\therefore Q=mC \Delta T=200\times 0.09\times 60$
$=1080 cal$.

If $20 \ g$ of ice at $0^\circ C$ is mixed with $10 \ g$ of water at $40^\circ C$, the final mass of water in the mixture is:

  1. 10 g

  2. 25 g

  3. 18 g

  4. 20 g


Correct Option: A

A liquid P if specific heat capacity $2400 J kg^{-1} K^{-1}$ and at $70^oC$ is mixed with another liquid R of specific heat capacity $1000 J kg^{-1} K^{-1}$ at $30^oC$. After mixing, the final temperature of the mixture is $40^oC$. Find the ratio of the mass of the liquids mixed?

  1. 4 : 5

  2. 8 : 5

  3. 40 : 5

  4. 48 : 5


Correct Option: D
Explanation:

Let mass of the liquids P and R be $m _P$ and $m _R$ respectively.
According to calorimeter,
Heat lost by liquid P $=$ Heat gained by liquid R.
$\therefore m _PC _P(70-40)=m _RC _R(40-30)$
$m _P(2400)(40)=m _R(1000)(10)$
$\frac {m _P}{m _R}=\frac {2400\times 40}{1000\times 10}=\frac {48}{5}$
$\frac {m _P}{m _R}=48 : 5$

Calculate the amount of heat required to convert 5 kg of ice to $0^oC$ to vapour at $100^oC$.

  1. $1.5\times 10^7 J$

  2. $2.5\times 10^7 J$

  3. $3.5\times 10^7 J$

  4. $4.5\times 10^7 J$


Correct Option: A
Explanation:

$m=5 kg, L _f=3.36\times 10^5 J kg^{-1}$
$L _V=2.25\times 10^6 J kg^{-1}$
$C=4200 J kg^{-1} {\;}^oC^{-1}$ Then $Q=?$
$Q=$ Heat required to convert ice at $^oC$ to water $0^oC +$ Heat required to convert water at $100^oC$ to vapour at $100^oC$
$=mL _f+ms(100-0)+mL$
$=5\times 3.36\times 10^5+5\times 4200\times (100-0)+5\times 2.25\times 10^6$
$=16.8\times 10^5+21\times 10^5+112.5\times 10^5$
$=1.5\times 10^7J$

In a calorimeter of water equivalent $20g$,water of mass $1.1$kg  is taken at $288K$ temperature.If steam at temperature $373K$ is passed through it and temperature of water increases by $6.5^oC$ then the mass of steam condensed is:

  1. $17.5g$

  2. $11.7g$

  3. $15.7g$

  4. $18.2g$


Correct Option: B

Steam at $100^oC$ is passed into $2.0$kg of water contained in a calorimeter of water equivalent $0.02$kg at $15^oC$ till the temperature of the calorimeter and its content rise to $90^oC$. The mass of steam condensed in kg is

  1. $0.301$

  2. $0.280$

  3. $0.60$

  4. $0.02$


Correct Option: B

Utensils used for efficient cooking should have

  1. Large heat capacity

  2. Small heat capacity

  3. Medium heat capacity

  4. Any heat capacity


Correct Option: A

A copper calorimeter of a mass $300\ g$ contains $500\ g$ of water at a temperature of $20^\circ C$. A $500\ g$ of copper block at $100^\circ C$ is dropped into the calorimeter. If the resultant temperature is $25^\circ C$, then fond the specific heat of copper in $JKg^{-1} K^{-1}$.

  1. $190$

  2. $290$

  3. $390$

  4. $490$


Correct Option: C

Calories is defined as the amount of heat required to rise temperature of $1\ g$ of water by $1^{o}C$ and it is defined under which of the following conditions.

  1. From $14.5^{o}C$ to $15.5^{o}C$ at $760\ mm$ of $Hg$

  2. From $98.5^{o}C$ to $99.5^{o}C$ at $760\ mm$ of $Hg$

  3. From $13.5^{o}C$ to $14.5^{o}C$ at $76\ mm$ of $Hg$

  4. From $3.5^{o}C$ to $4.5^{o}C$ at $76\ mm$ of $Hg$


Correct Option: A

A copper calorimeter contains $100 g$ of water at $16^o C$. When $15 g$ of ice is added to it, the resultant temperature of the mixture is $4^o C$. Water equivalent of the calorimeter is

  1. $8 g$

  2. $12 g$

  3. $6 g$

  4. None


Correct Option: D

 Which of the following material is used to make calorimeter?

  1. Glass

  2. Ebonite

  3. Metal

  4. Superconductor


Correct Option: C

The water equivalent of a 400 g copper calorimeter (specific heat =0.1 cal/$g^{ o }C$)-

  1. $40g$

  2. $4000g$

  3. $200g$

  4. $4g$


Correct Option: A

A man would feel iron and wooden balls equally cold or hot at

  1. $98.6^oC$

  2. $98.6^oF$

  3. $198.6^oF$

  4. $198.6^oC$


Correct Option: B
Explanation:

A man would find them equally cold or hot only when the heat flowing in or out of them is equal.
But since there would be the same temperature difference between the object and the man.
The heat flow would only be the same, when there is no heat flow. i.e. the body's temperature should be equal to the body temperature of the man. Which is $98.6 ^{\circ} F$

How many calories of heat are required by gram of water at $99^oC$ to boil off:

  1. 530

  2. 640

  3. 540

  4. 500


Correct Option: C
Explanation:

Heat $= 4.2(1)(1)+ 1(2260) = 2264.2 J$
$1 cal = 4.2J$
Thus, Heat $= 539.1 cal \approx 540 cal$

$50  g$ of ice at 0 C is mixed with $50  g$ of water at 20 C.The resultant temperature of the mixture would be

  1. 10 C

  2. 0 C

  3. -10 C

  4. -35 C


Correct Option: B
Explanation:

$50 g $ of Ice at $0^0C$ has a latent heat of

$ Q = m \times L = 50 \times 80 $ 

                    $ = 4000 cal $

Now for water to reach $0^{0}C$ without changing its state 

Heat released by water = $ mC _p  \Delta T$

                                      = $ 50 \times 1 \times 20  $

                                      = $1000 cal $

As the heat to be removed from water is less than the latent heat of $50g $ of ice, the resultant mixture stays at $0^0C $ temperature.

At which temperature do the readings of the celcius and the Fahrenheit scales coincide ?

  1. $0$

  2. $100$

  3. $-40$

  4. $-80$

  5. None of the above


Correct Option: C
Explanation:

We know the relation between celsius and Fahrenheit is given by , 

$ ^0 C = \dfrac{9}{5}  \times ^0F + 32 $
For readings to coincide ,  $ ^0 C = ^0 F$
$ ^0 C = \dfrac{9}{5}  \times ^0C + 32 $
On Solving, $ ^0 C = -40 $


The amount of heat required to convert 1 g of ice (specific 0.5 cal  at $g^{-1o} C^{-1}$ ) at $-10^0 C$ to steam at $100 $ $^\circ C$ is ___________.

[ Given: Latent heat of ice is $80 Cal/ gm,$ Latent heat of steam is $540 Cal/gm $, Specific heat of water is $1 Cal/gm/C$ ]

  1. 725 cal

  2. 636 cal

  3. 716 cal

  4. None of these


Correct Option: A
Explanation:

Amount of heat required = Change the temp Of Ice from -10 C to 0 C + Heat required to melt the Ice + Heat required to increase the temperature of water from 0 to 100 C + Heat required to convert water  at 100 C to vapor at 100 C

$=1\times 0.5[0-(-10)]+1\times 80+1\times 1\times 100+1\times 540\ =5+80+100+540\ =725cal  $

The specific heat for substance $A$ is twice the specific heat of substance $B$. The same mass of each substance is allowed to gain $50$ Joules of heat energy. As a result of the heating process:

  1. the temperature of $A$ rises twice as much as $B$

  2. the temperature of $A$ rises four times as much as $B$

  3. the temperature of $B$ rises twice as much as $A$

  4. the temperature of $B$ rises four times as much as $A$

  5. the temperature of both $B$ and $A$ rise the same amount


Correct Option: C
Explanation:

Let specific heat of substance $A$  is $2c$ and specific heat of substance $B$ is $c$ , 

we have ,  heat given  $Q=mc\Delta t$, where  $\Delta t $ denotes change in temperature , 
so , for substance $A$, $Q=m\times 2c\Delta t _{A}$ 
or $\Delta t _{A}=Q/2mc$ .........eq1
For substance $B$ ,   $Q=mc\Delta t _{B}$   
$\Delta t _{B}=Q/mc$ ...................eq2
by eq1 and eq2,
$2\Delta t _{A}=\Delta t _{B}$     

To measure the specific heat of copper, an experiment is performed in the lab. A piece of copper is heated in an oven then dropped into a beaker of water. To calculate the specific heat of copper, the experimenter must know or measure the value of all of the quantities below EXCEPT the

  1. Original temperatures of the copper and the water

  2. Mass of the water

  3. Final (equilibrium) temperature of the copper and the water

  4. Time taken to achieve equilibrium after the copper is dropped into the water

  5. Specific heat of the water


Correct Option: B
Explanation:

Specific of  a substance  is given by: $\Delta Q$= $mc\Delta T$

where, $\Delta Q=$ heat given to substance
                $m=$ mass  of  the substance
             $\Delta T=$ increase in temperature of substance (for that we require initial and final temperature of substance)
If the substance is copper in this experiment the experimenter requires the mass of copper piece not the mass of water because water is just dropping the temperature of copper  piece not more  than this.        

An aluminium block of 2m mass and an iron block of m mass,each absorbs the same amount of heat, and both blocks remain solid. If the specific heat of aluminium is twice the specific heat of iron, then find out the correct statement?

  1. The increase in temperature of the aluminum block is twice the increase in temperature of the iron block

  2. The increase in temperature of the aluminum block is four times the increase in temperature of the iron block

  3. The increase in temperature of the aluminum block is the same as increase in temperature of the iron block

  4. The increase in temperature of the iron block is twice the increase in temperature of the aluminum block

  5. The increase in temperature of the iron block is four times the increase in temperature of the aluminum block


Correct Option: E
Explanation:

The heat required to rise the temperature of body of mass $m$ of specific heat $s$ by $\Delta T=H=ms\Delta T$

Thus for same amount of heat, the rise in temperature ration of aluminium and iron is $\dfrac{\Delta T _{Al}}{\Delta T _{Fe}}=\dfrac{m _{iron}s _{iron}}{m _{aluminium}s _{aluminium}}=\dfrac{1}{4}$
Thus the rise in temperature of the iron block is four times the increase in temperature of the aluminum block

A mass of stainless steel spoon is 0.04 kg and specific heat is $0.50 kJ/kg \times ^oC$. Then calculate the heat which is required to raise the temperature $20^oC$ to $50^oC$ of the spoon.

  1. 200 J

  2. 400 J

  3. 600 J

  4. 800 J

  5. 1,000 J


Correct Option: C
Explanation:

The required heat, $Q=ms\Delta T=0.04\times (0.50\times 10^3)\times(50-20)=600 J$

A body having $1680 J$ of energy is supplied to $100 g$ of water. If the entire amount of energy is converted into heat the rise in temperature of water (sp. heat of water = $4200 JKg^{ -1 }\ ^0C ^{ -1 } $)

  1. $0.4^{ \circ }{ C }$

  2. $40^{ \circ }{ C }$

  3. $4^{ \circ }{ C }$

  4. $44^{ \circ }{ C }$


Correct Option: C
Explanation:

If $\delta T$ is the rise in temperature, then the amount of heat supplied is $Q=mS\Delta T$  where $S=$ specific heat

Thus, $1680=(100/1000)(4200)\Delta T$ or $\Delta T=4^oC$

$5gm$ of steam at $100^oC$ is passed into calorimeter containing liquid , Temperature of liquid rises from $32^oC$ to $40^oC$. Then water equivalent of calorimeter and content is 

  1. $40$ gram

  2. $375$ gram

  3. $300$ gram

  4. $160$ gram


Correct Option: B
Explanation:

Latent heat of vaporization of water = $2260kJ/Kg$

The specific heat capacity of water = $4185.5 J/Kg$

Heat lost by steam = heat gained by the water and calorimeter

Formula :

Heat gained by water = mcФ

m = mass of water

c = specific heat capacity of water.

$Ф = Change in temperature.  = 40 - 32 = 8$

Heat lost by steam = mLv + mcФ

Lv = latent heat of vaporisation

m = mass of steam.

$Ф = 100 - 40 = 60$

Doing the substitution :

$2260000 \times 0.005 + 60 \times 0.005 \times 4185.5 = m \times 4185.5 \times 8$

$12555.65 = 33484m$

$m = \dfrac {12555.65}  {33484} = 0.37497 kg$

$= 0.37497 \times 1000 = 374.97 kg$

$= 374.97kg$

1 kg of water at $20^{\circ}C$ is, mixed with 800 g of water at $80^{\circ}C$. Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.

  1. $24.44^{\circ} C$

  2. $46.67^{\circ} C$

  3. $44.44^{\circ} C$

  4. $54.44^{\circ} C$


Correct Option: B
Explanation:

$mc\theta _1 + mc\theta _2 = mc\theta _0$
since c is constant, assuming water is heated from $0^o C$
(1000)(20) + (800)(80)= (1800)$\theta _0$
20000+64000= 84000=1800$\theta _0$
$\theta _0= 46.67^o C$
hence the finale temperature in $\theta _0=46.67^o$

The temperature of equal masses of three different liquids A, B, and C are $12^o C$,$19^o C$ and $28^o C$ respectively. The temperature when A and B are mixed is $16^oC$ and When B and C are mixed is $23^o C$. The temperature when A and C are mixed is:

  1. $18.2^ C$

  2. $22^ C$

  3. $20.3^ C$

  4. $24.2^ C$


Correct Option: C
Explanation:

let $m _1=m _2=m _3=m$
Let $s _1,s _2,s _3$ be the respective specific heats of the three liquids.
When A and B are mixed, temperature of mixed = $16^o C$
A heat gained by A = heat lost by B
$\therefore ms _1(16-12)=ms _2(19-16)$
$4s _1=3s _2$.......(i)
When B and C are mixed , temperature of mixture$=23^o C$.
As heat gained by B = heat lost by C,
$ms _2(23-19)=ms _3(28-23)$
$\therefore 4s _2=5s _3$......(ii)
From (i) and (ii) , $=\dfrac{3}{4}s _2=\dfrac{15}{16}s _3$
When A and C are mixed, suppose temperature of mixture$=t$
Heat gained by A = Heat lost by C
$ms _1(t-12)=ms _3(28-t)$
$\dfrac{15}{16}s _2(t-12)=s _3(28-t)$
$15t-180=448-16t$
$31t=448+180=628$
$t=\dfrac{628}{3}=20.3^o C$

An adulterated sample of milk has a density, 1032 kg m$^{-3}$, while pure milk has a density of 1080 kg m$^{-3}$. Then the volume of pure milk in a sampled of 10 litres of adulterated milk is:

  1. 1 litre

  2. 2 litre

  3. 3 litre

  4. 4 litre


Correct Option: D
Explanation:

Mass of adulterated milk
$M _A = 1032 \times (10 \times 10^{-3}$) kg
or $M _A = 10.32 kg$      ($\because 1 litre = 10^{-3} m^3$)
$\therefore$ Mass of pure milk $M _p= 1080 \times V _p$ 

$\therefore$ Mass of water added = $\rho _wV _w= M _A - M _p$
$\therefore$ 10$^3 \times$ (volume of water added)= $M _A - M _p$
$\therefore10^3\times(10 \times 10^{-3}- V _p) = 10.32 - 1080 V _p$
or 80 $V _p = 0.32$
or $V _p$ =  $\displaystyle \frac{0.32}{80}$
$= \dfrac{1}{250} m^3=\dfrac{1000}{250} litre = 4 litre$.

An experiment requires a gas with $\gamma = 1.50$. This can be achieved by mixing together monatomic and rigid diatomic ideal gases. The ratio of moles of the monatomic to diatomic gas in the mixture is

  1. $1 : 3$

  2. $2 : 3$

  3. $1 : 1$

  4. $3 : 4$


Correct Option: C
Explanation:

One mole of an ideal monoatomic gas is is C$ _{v}$ = $\dfrac{3}{2}$R and C$ _{p}$ = $\dfrac{5}{2}$R


i.e $\gamma$ = 1.66 for monoatomic gas

For One mole of an ideal dioatomic gas,
$\gamma$ = 1.4 for air which is pre dominantly a  diatomic gas
If we take 1 mole monoatomic and 1 mole of diatomic gas in a mixture then we get the following result;

$\gamma$ = $\dfrac{n1\gamma + n2\gamma}{n1 + n2}$ 

Now since we have taken the no. of moles of monoatomic as well as diatomic as 1, therefore
$\gamma$ = $\dfrac{y1 + y2}{2}$ where $\gamma$1 and $\gamma$2 are the values of $\dfrac{C _p}{C _v}$ for individual gases.

Substuting the values of C$ _p$ and C$ _v$ i.e $\gamma$1 = 1.6 and $\gamma$2 = 1.4 we get
$\gamma$ = 1.53 which is approximately equal to 1.50 which is given.
Hence by taking 1 mole og monoatomic and 1 mole of diatomic mixture we got $\gamma$ as 1.50
Hence the ratio of moles of monoatomic to diatomic gas in the mixture is 1:1

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