A Carnot engine work as refrigerator in between `0^(@)C` and `27^(@)C`. How much energy is needed to freeze `10 kg` ice at `0^(@)C`/

Answer

To solve the problem of how much energy is needed to freeze 10 kg of ice at 0°C using a Carnot refrigerator operating between 0°C and 27°C, we will follow these steps:

### Step 1: Convert Temperatures to Kelvin
The temperatures given are:
- T1 = 27°C
- T2 = 0°C

To convert these to Kelvin:
- T1 = 27 + 273 = 300 K
- T2 = 0 + 273 = 273 K

### Step 2: Calculate the Coefficient of Performance (β)
The coefficient of performance (β) for a Carnot refrigerator is given by the formula:
\[ \beta = \frac{T2}{T1 - T2} \]
Substituting the values:
\[ \beta = \frac{273}{300 - 273} = \frac{273}{27} = 10.11 \]

### Step 3: Calculate the Energy Required to Freeze Ice
The energy required to freeze ice can be calculated using the formula:
\[ Q = m \cdot L_f \]
where:
- \( m = 10 \, \text{kg} = 10,000 \, \text{grams} \)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion for ice)

Calculating \( Q \):
\[ Q = 10,000 \, \text{g} \cdot 80 \, \text{cal/g} = 800,000 \, \text{cal} \]

### Step 4: Relate Work Done (W) to Heat Absorbed (Q)
The work done (W) by the refrigerator can be calculated using the relationship:
\[ W = \frac{Q}{\beta} \]
Substituting the values:
\[ W = \frac{800,000 \, \text{cal}}{10.11} \approx 79,12 \, \text{kcal} \]

### Final Answer
The energy needed to freeze 10 kg of ice at 0°C is approximately:
\[ W \approx 79.12 \, \text{kcal} \]

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