Tag: measuring thermal quantities by the method of mixtures

Questions Related to measuring thermal quantities by the method of mixtures

A thermometer is used to measure 

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

  1. heat

  2. thermal capacity

  3. water equivalent

  4. temperature

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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.

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

  1. I only

  2. I and II only

  3. I, II, and III

  4. II only

  5. I and III only

Show answer
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 ?

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

  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

Show answer
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.

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