Tag: calorimetry

Questions Related to calorimetry

$5$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,

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

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 $2^oC$,
$W= (10000)(2)(2)= 40000 J$

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 .

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 $Hg$.

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
$To=75^o C$
$\Rightarrow \ 100\times 0.1\times (75-T)+100\times 0.1(75-30)+1.70 \times 1\times (75.32)$
$\therefore \ T=885^oC$

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

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

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

$50  g$ 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.