Tag: calorimeter

Questions Related to calorimeter

How much heat is required to raise the temperature of 150 g of iron from $20 ^oC$ to $25 ^oC$? (Specific heat of iron $480 J kg^{-1} {\;}^oC^{-1})$

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 350 J

  2. 345 J

  3. 360 J

  4. 330 J

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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 $=390 J kg^{-1} {\;}^oC^{-1})$.

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 50000 J

  2. 10000 J

  3. 10920 J

  4. 10900 J

Show answer
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 $=4200 J kg^{-1} {\;}^oC^{-1})$

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. $0.4 ^oC$

  2. $40 ^oC$

  3. $4 ^oC$

  4. $44 ^oC$

Show answer
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?

physics thermal properties heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. Calorimetry

  2. Fermentation

  3. Latent heat

  4. Hidden heat

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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 $95 ^oC$?

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 900 kcal

  2. 90 kcal

  3. 10 kcal

  4. 9 kcal

Show answer
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.

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. True

  2. False

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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?

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. $45^oC$

  2. $65^oC$

  3. $85^oC$

  4. $35^oC$

Show answer
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 $=80 cal g^{-1}$.

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. $0^oC$

  2. $1^oC$

  3. $10^oC$

  4. $20^oC$

Show answer
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 :

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 15 cal

  2. 25 cal

  3. 50 cal

  4. 100 cal

Show answer
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.

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 253 K

  2. 260 K

  3. 273 K

  4. 290 K

Show answer
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.