Question

Heat required in converting 1g of ice at $-10{}^{\circ }C$ into steam at $100{}^{\circ }C$ is:
Latent heat of fusion = 80 cal/g
Latent heat of vaporization = 540 cal/g.
A. 3045 J
B. 6056 J
C. 721 J
D. 6 J

Answer 1

Divide the entire process into 4 steps. In the steps where the temperature of the substance is to be raised use $Q=mc\Delta T$ and when the substance changes its phase, use $Q=mL$. Then add the heat required in all the four steps.

Formula used: $Q=mc\Delta T$
$Q=mL$

Complete step by step answer:
It is given that there is 1 gram of ice at a temperature of -10$^{\circ }$C. It is told to convert this 1 gram of ice at -10$^{\circ }$C into steam at 100$^{\circ }$C. This means that we have to supply a specific amount of heat to the ice in order to raise its temperature.
The process of converting the given mass of ice to steam involves four steps.

1. Raising the temperature of the 1 g ice from -10$^{\circ }$C to 0$^{\circ }$C.
2. Conversion of ice to water at 0$^{\circ }$C.
3. Raising the temperature of the 1 g water from 0$^{\circ }$C to 100$^{\circ }$C.
4. Conversion of water to steam at 100$^{\circ }$C.

All the four steps require heat. Let us calculate the heat required in each of the steps.

1. The heat required to raise the temperature of a substance of mass m by an amount $\Delta$T is given as $Q=mc\Delta T$, where c is the specific heat capacity of the substance.
   The specific heat capacity of ice is $0.5cal{{g}^{-1}}{}^{\circ }{{C}^{-1}}$.
   Let the heat required to raise the temperature of ice from -10$^{\circ }$C to 0$^{\circ }$C be ${{Q}_{1}}$. Here m=1g and $\Delta T=10{}^{\circ }C$
   Hence, ${{Q}_{1}}=1\times 0.5\times 10=5cal$.
2. When ice converts into water, the heat required is equal to $Q=m{{L}_{f}}$, where ${{L}_{f}}$ is the latent heat of fusion.
   It is given that ${{L}_{f}}$= 80cal/g. Let the heat required by the conversion be ${{Q}_{2}}$.
   Hence, ${{Q}_{2}}=1\times 80=80cal$.
3. The ice is completely converted into water and now we are raising the temperature of 1g water from 0$^{\circ }$C to 100$^{\circ }$C. Hence, $\Delta T=100{}^{\circ }C$. The specific heat capacity of water is $0.5cal{{g}^{-1}}{}^{\circ }{{C}^{-1}}$. Let the heat required be ${{Q}_{3}}$.
   Hence, ${{Q}_{3}}=1\times 1\times 100=100cal$.
4. When water at 100$^{\circ }$C converts into steam, the heat required is equal to $Q=m{{L}_{v}}$, where ${{L}_{v}}$ is the latent heat of vaporization.
   It is given that ${{L}_{v}}$= 540cal/g. Let the heat required by the conversion be ${{Q}_{4}}$.
   Hence, ${{Q}_{4}}=1\times 540=540cal$.
   Therefore, the total heat required for the whole process is $Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}$
   $\Rightarrow Q=5+80+100+540=725cal$
   And 1cal=4.2 J
   Therefore,
   $\Rightarrow Q=725cal=725\times 4.2J=3045J$

So, the correct answer is “Option A”.

Note: Students may make a mistake by taking the value of specific heat capacity for ice and water as 1cal/g/K.
Always remember that when the temperature of a substance is to be raised, heat is supplied to the substance and when the temperature of a substance is to be lowered, heat is taken out from the substance.