Calculate the least amount of work that must be done to freeze one gram of wate at `0^@C` by means of a refrigerator. Temperature of surroundings is `27^@C`. How much heat is passed on the surroundings in this process? Latent heat of fusion `L=80cal//g`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We need to calculate the least amount of work required to freeze 1 gram of water at 0°C using a refrigerator, with the surrounding temperature at 27°C. We also need to find out how much heat is rejected to the surroundings in this process.

### Step 2: Identify the given data
- Mass of water (m) = 1 gram
- Latent heat of fusion (L) = 80 cal/g
- Temperature of surroundings (T_high) = 27°C = 300 K
- Temperature of water (T_low) = 0°C = 273 K

### Step 3: Write the formula for Coefficient of Performance (COP)
The Coefficient of Performance (COP) for a refrigerator is given by:
\[ COP = \frac{Q_2}{W} \]
Where:
- \(Q_2\) is the heat absorbed from the cold reservoir (water).
- \(W\) is the work done by the refrigerator.

### Step 4: Express \(Q_2\)
The heat absorbed by the refrigerator when freezing 1 gram of water is given by:
\[ Q_2 = m \cdot L \]
Substituting the values:
\[ Q_2 = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 5: Relate COP to temperatures
The COP can also be expressed in terms of temperatures:
\[ COP = \frac{T_{low}}{T_{high} - T_{low}} \]
Substituting the temperatures:
\[ COP = \frac{273}{300 - 273} = \frac{273}{27} = 10.11 \]

### Step 6: Relate COP to work done
From the COP formula, we can express work done (W):
\[ W = \frac{Q_2}{COP} \]
Substituting the values:
\[ W = \frac{80 \, \text{cal}}{10.11} \approx 7.91 \, \text{cal} \]

### Step 7: Calculate heat rejected to surroundings
The heat rejected to the surroundings (\(Q_1\)) can be calculated using:
\[ Q_1 = Q_2 + W \]
Substituting the values:
\[ Q_1 = 80 \, \text{cal} + 7.91 \, \text{cal} \approx 87.91 \, \text{cal} \]

### Final Results
- The least amount of work done to freeze 1 gram of water at 0°C is approximately **7.91 cal**.
- The heat passed to the surroundings is approximately **87.91 cal**.

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