Question

A refrigerator converts 100g of water at ${{20}^{0}}C$to ice at $-{{10}^{0}}C$ in 35 minutes. Calculate the average rate of heat extraction in terms of watts.
Given:
Specific heat capacity of ice $=2.1J{{g}^{-1}}^{0}{{C}^{-1}}$
Specific heat capacity of water $=4.2J{{g}^{-1}}^{0}{{C}^{-1}}$
Specific latent heat of fusion of ice $336J{{g}^{-1}}$

Answer 1

Hint: To solve this question first find the heat extracted from the water in converting to ice step by step using the concept of the specific heat capacity and the specific latent heat of fusion. Then find the average rate of heat extraction by dividing the total heat extracted by the total time required to extract the heat.

Complete answer:
The mass of the water is given as, $m=100g$
The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 gram of substance by one degree Celsius.
The heat required to change temperature of a substance can be expressed in terms of specific heat capacity of the substance as,
$Q=mc\Delta T$
Where, Q is the heat, m is the mass of the substance, c is the specific heat capacity of the substance and $\Delta T$is change in temperature of the substance.
Now, the refrigerator converts 100g of water at ${{20}^{0}}C$to ice at $-{{10}^{0}}C$ in 35 minutes.
Specific heat capacity of water $=4.2J{{g}^{-1}}^{0}{{C}^{-1}}$
Amount of heat extracted to change the temperature of water from ${{20}^{0}}C$to ice at ${{0}^{0}}C$ is,
$\begin{align}
  & {{Q}_{1}}=m{{c}_{w}}\Delta T=100\times 4.2\times \left( 20-0 \right) \\
 & {{Q}_{1}}=8400J \\
\end{align}$
Latent heat is the heat required per unit mass to change the phase of the substance.
Specific latent heat of fusion of ice $336J{{g}^{-1}}$
Amount of heat required to convert water to ice at ${{0}^{0}}C$ is,
\[\begin{align}
  & {{Q}_{2}}=mL=100\times 336 \\
 & {{Q}_{2}}=33600J \\
\end{align}\]
Specific heat capacity of ice $=2.1J{{g}^{-1}}^{0}{{C}^{-1}}$
Amount of heat extracted to change the temperature of ice from ${{0}^{0}}C$to ice at $-{{10}^{0}}C$ is,
$\begin{align}
  & {{Q}_{3}}=m{{c}_{i}}\Delta T=100\times 2.1\times \left( 0-\left( -10 \right) \right) \\
 & {{Q}_{3}}=2100J \\
\end{align}$
Total heat extracted is,
$\begin{align}
  & Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}} \\
 & Q=8400J+33600J+2100J \\
 & Q=44100J \\
\end{align}$
The heat is extracted over a time of 35 minutes. So, the average rate of heat extraction is,
$P=\dfrac{Q}{t}=\dfrac{44100J}{35\times 60s}=21W$
So, the average rate of heat extraction is 21 watts.

Note:
The average rate of heat extraction can also be defined as the power of the refrigerator. The power is defined as the rate of amount of heat extracted from or given to the substance with respect to time required to extract or give the heat to the substance.