Tag: measurement and effects of heat

Questions Related to measurement and effects of heat

State True or False.


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

Show answer
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.

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

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

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

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

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

Show answer
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 ${ 25 }^{ \circ  }C$?

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. $330 J$

Show answer
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 ${ 95 }^{ \circ  }C$?

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:

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 ${ { g }^{ -1 } }^{ \circ  }{ C }^{ -1 }$)

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

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

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

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

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

Show answer
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 ${ 0 }^{ \circ  }C$?

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

  1. $400 cal$

  2. $4000 cal$

  3. $3000 cal$

  4. $300 cal$

Show answer
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 ${ 15 }^{ \circ  }C$. (${ L } _{ ice }\quad =\quad 3.34\quad \times \quad { 10 }^{ 5 }\quad J{ \quad kg }^{ -1 }$, ${ C } _{ water }\quad =\quad 4180\quad J{ \quad kg }^{ -1 }\quad ^{ \circ  }{ { C }^{ -1 } }$)

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

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

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

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

Show answer
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.

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 $10^oC$.