Question

What is the amount of heat (in calories) required to convert 10 g of ice at $-{{10}^{{}^\circ }}C$ into steam at ${{100}^{{}^\circ }}C$?
(Latent heat of vaporization is 536 cal/gm-K)
A). 5400 cal
B). 6400 cal
C). 7210 cal
D). 8250 cal

Answer 1

Hint: To convert a unit mass of a substance from one state to another (viz. Solid, liquid gas), heat is required which is known as latent heat. To increase the temperature of the unit mass of a substance by unity, the heat required is known as the specific heat of that substance. We will calculate the heat required for converting ice to steam by summing up the heat required in all these processes.

Formula used:
Heat during temperature change, $\Delta Q=ms\Delta T$ and latent heat, $\Delta Q=mL$

Complete step-by-step solution:
When ice at $-{{10}^{{}^\circ }}C$ is converted to steam at ${{100}^{{}^\circ }}C$, it undergoes following thermodynamic processes:
Conversion of ice at $-{{10}^{{}^\circ }}C$ to ice at ${{0}^{{}^\circ }}C$
Conversion of ice at ${{0}^{{}^\circ }}C$ to water at ${{0}^{{}^\circ }}C$
Changing temperature of water at ${{0}^{{}^\circ }}C$ to ${{100}^{{}^\circ }}C$
Conversion of water at ${{100}^{{}^\circ }}C$ to steam at ${{100}^{{}^\circ }}C$
The heat required for converting the ice at $-{{10}^{{}^\circ }}C$ to steam at ${{100}^{{}^\circ }}C$ is the sum of heat required in each process.
The temperature of a substance remains constant when its state is changing. The heat required by per unit mass of the substance to change its state is known as latent heat. The heat required by the substance to convert from one state to another is given by
$\Delta Q=mL$ where $m$ is the mass of the substance and $L$ is latent heat.
The heat required to change the temperature of a substance is given by
$\Delta Q=ms\Delta T$ where $s$ is the specific heat of the substance and $\Delta T$ is the change in temperature.
Therefore, the total heat required to convert ice at $-{{10}^{{}^\circ }}C$ to steam at ${{100}^{{}^\circ }}C$ is given as
$\Delta {{Q}_{total}}=\Delta {{Q}_{1}}+\Delta {{Q}_{2}}+\Delta {{Q}_{3}}+\Delta {{Q}_{4}}$
The number at subscript denotes the number of thermodynamic processes.
$\Delta {{Q}_{total}}=m{{s}_{ice}}\Delta {{T}_{1}}+m{{L}_{fusion}}+m{{s}_{water}}\Delta {{T}_{2}}+m{{L}_{vaporisation}}$
According to the question,
$m=10g$
$\Delta {{T}_{1}}={{10}^{{}^\circ }}C$ (ice at $-{{10}^{{}^\circ }}C$ to ice at ${{0}^{{}^\circ }}C$)
$\Delta {{T}_{2}}={{100}^{{}^\circ }}C$ (water at ${{0}^{{}^\circ }}C$ to ${{100}^{{}^\circ }}C$)
${{L}_{fusion}}=80\,cal/g$
${{L}_{vaporisation}}=536\,cal/g$
${{s}_{ice}}=0.5\,cal/{{g}^{{}^\circ }}C$
${{s}_{water}}=1\,cal/{{g}^{{}^\circ }}C$
Substituting these values, we get
$\Delta {{Q}_{total}}=10[(0.5\times 10)+80+(1\times 100)+536]$
$\Delta {{Q}_{total}}=7210\,cal$

Hence, option C is correct.

Note: Temperature of the substance does not change when the state of substance changes. When water is vaporized at ${{100}^{{}^\circ }}C$, its temperature does not change but its internal energy increases by a magnitude equivalent to heat it absorbs during the change of state.