Tag: calorimeter

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Questions Related to calorimeter

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 .

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

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 ?

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

State True or False.


According to principle of calorimetry heat absorbed by cold bodies is equal to heat released by hot bodies.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

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.

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

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

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

  4. ${ 43.2 }^{ \circ }C$
Reveal answer Fill a bubble to check yourself
B Correct answer
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                 $\theta _{2}=100-59.8=40.2^{o}C$

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

How much heat is required to raise the temperature of $150 g$ of iron from ${ 20 }^{ \circ  }C$ to ${ 25 }^{ \circ  }C$?

  1. $350 J$

  2. $345 J$

  3. $360 J$

  4. $330 J$
Reveal answer Fill a bubble to check yourself
B Correct answer
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                 $Q=150\times0.46\times(25-20)=345J$

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

How much heat is required to raise the temperature of $100 g$ of water from ${ 5 }^{ \circ  }C$ to ${ 95 }^{ \circ  }C$?

  1. $900 kcal$

  2. $90 kcal$

  3. $10 kcal$

  4. $9 kcal$
Reveal answer Fill a bubble to check yourself
D Correct answer
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                 $Q=100\times1\times(95-5)=9000cal=9kcal$

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

2000 cal of heat is supplied to 200 g of water. Find the rise in temperature. (Specific heat of water = 1 cal ${ { g }^{ -1 } }^{ \circ  }{ C }^{ -1 }$)

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

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

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

  4. ${ 40 }^{ \circ }C$
Reveal answer Fill a bubble to check yourself
A Correct answer
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                 $\Delta \theta=Q/(mc)=2000/(200\times1)=10^{o}C$

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

What will be the amount of heat required to convert $50 g$ of ice at ${ 0 }^{ \circ  }C$ to water at ${ 0 }^{ \circ  }C$?

  1. $400 cal$

  2. $4000 cal$

  3. $3000 cal$

  4. $300 cal$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 ,           $Q=50\times80=4000cal$ 

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

Calculate the quantity of heat required to convert 1.5 kg of ice at ${ 100 }^{ \circ  }C$ to water at ${ 15 }^{ \circ  }C$. (${ L } _{ ice }\quad =\quad 3.34\quad \times \quad { 10 }^{ 5 }\quad J{ \quad kg }^{ -1 }$, ${ C } _{ water }\quad =\quad 4180\quad J{ \quad kg }^{ -1 }\quad ^{ \circ  }{ { C }^{ -1 } }$)

  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. $4.95\quad \times \quad { 10 }^{ 5 }\quad J$
Reveal answer Fill a bubble to check yourself
B Correct answer
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          $Q=5.01\times10^{5}+0.94\times10^{5}=5.95\times10^{5}J$