Tag: heat exchange

Questions Related to heat exchange

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?

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

  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

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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.

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:
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 $)

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

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

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

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

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

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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.

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

  1. $30.2^oC$

  2. $40.2^oC$

  3. $50.2^oC$

  4. $43.2^oC$

Show answer
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 $=1 cal g^{-1} {\;}^oC^{-1})$

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

  1. $10 ^oC$

  2. $20 ^oC$

  3. $30 ^oC$

  4. $40 ^oC$

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

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

  1. 2400 J

  2. 5000 J

  3. 40000 J

  4. 42000 J

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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 $=4.2 J g^{-1} {\;}^oC^{-1}$.

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

  1. 25.5 kJ

  2. 29.4 kJ

  3. 30 kJ

  4. 40 kJ

Show answer
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 $=0.09 cal g^{-1} {\;}^oC^{-1})$?

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

  1. 1080 cal

  2. 1000 cal

  3. 1500 cal

  4. 1100 cal

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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:

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

  1. 10 g

  2. 25 g

  3. 18 g

  4. 20 g

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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?

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

  1. 4 : 5

  2. 8 : 5

  3. 40 : 5

  4. 48 : 5

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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.
$\therefore m _PC _P(70-40)=m _RC _R(40-30)$
$m _P(2400)(40)=m _R(1000)(10)$
$\frac {m _P}{m _R}=\frac {2400\times 40}{1000\times 10}=\frac {48}{5}$
$\frac {m _P}{m _R}=48 : 5$