Question

Suppose a refrigerator does minimum possible work in converting 100 grams of water at 0°C to ice. How much heat (in calories) is released to the surroundings at a temperature of 27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer?

Answer 1

Solution:

Latent heat of fusion:

• Latent heat of fusion, also known as enthalpy of fusion, is the amount of energy that must be supplied to a solid substance (typically in the form of heat) in order to trigger a change in its physical state and convert it into a liquid (when the pressure of the environment is kept constant).
• For example, the latent heat of fusion of one kilogram of water, which is the amount of heat energy that must be supplied to convert 1 kg of ice without changing the temperature of the environment (which is kept at zero degrees celsius) is 333.55 kilojoules.

$\mathrm{Q}=\mathrm{mL}=8000\mathrm{cal}$

$$\begin{gathered} {\mathrm{Q}}_2=\mathrm{W}+{\mathrm{Q}}_1 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{performance}\left(\mathrm{COP}\right)=\frac{{\mathrm{Q}}_1}{\mathrm{W}}=\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2-{\mathrm{Q}}_1}=\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2-{\mathrm{T}}_1} \\ \frac{{\mathrm{Q}}_1}{\mathrm{W}}=\frac{273}{300-273} \\ \frac{{\mathrm{Q}}_1}{\mathrm{W}}=\frac{273}{27} \\ \mathrm{W}=\frac{27}{273}{\mathrm{Q}}_1 \\ \mathrm{W}=\frac{27}{273}\times 80\times 100 \\ {\mathrm{Q}}_2=\frac{27}{273}\times 80\times 100+80\times 100 \\ {\mathrm{Q}}_2=791.208+8000 \\ {\mathrm{Q}}_2=8791.208\mathrm{cal} \end{gathered}$$

The amount of heat released into the surroundings at a temperature of 27°C is 8791 calories.