A Carnot refrigerator extracts heat from water at `0^@C` and rejects it to room at `24.4^@C`. The work required by the refrigerator for every 1 kg of water converted into ice is
(Latent heat of ice=`336 kJ kg^(-1)`) a) 24.4kJ b) 30 kJ c) 336 kJ d) 27.55 kJ

Options

A. 24.4kJ
B. 30 kJ
C. 336 kJ
D. 11.2 kJ

Answer

To solve the problem of how much work is required by the Carnot refrigerator for converting 1 kg of water into ice, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
- The temperature of the cold reservoir (TL) is given as \(0^\circ C\).
- To convert to Kelvin:
\[ T_L = 0 + 273 = 273 \, K \]
- The temperature of the hot reservoir (TH) is given as \(24.4^\circ C\).
- To convert to Kelvin:
\[ T_H = 24.4 + 273 = 297.4 \, K \]

### Step 2: Calculate the efficiency of the Carnot refrigerator
- The efficiency (β) of a Carnot refrigerator is given by the formula:
\[ \beta = \frac{T_L}{T_H - T_L} \]
- Substituting the values:
\[ \beta = \frac{273}{297.4 - 273} = \frac{273}{24.4} \approx 11.2 \]

### Step 3: Use the efficiency to find the work done
- The work done (W) by the refrigerator can be calculated using the relationship:
\[ \beta = \frac{Q_H}{W} \]
- Rearranging gives:
\[ W = \frac{Q_H}{\beta} \]
- The latent heat (Q_H) for converting 1 kg of water into ice is given as \(336 \, kJ\).
- Substituting the values:
\[ W = \frac{336 \, kJ}{11.2} \approx 30 \, kJ \]

### Step 4: Conclusion
- The work required by the refrigerator for every 1 kg of water converted into ice is approximately \(30 \, kJ\).

Thus, the correct answer is **b) 30 kJ**.

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