In a refrigerator, the low temperature coil of evaporator is at `-23^@C` and the compressed gas in the condenser has a temperature of `77^@C`. The amount of electrical energy spent in freezing 1 kg of water at `0^@C` is

Options

A. `134400 J`
B. `1344 J`
C. `80000 J`
D. `3200 J`

Answer

To solve the problem of finding the amount of electrical energy spent in freezing 1 kg of water at \(0^\circ C\) in a refrigerator, we can follow these steps:

### Step 1: Convert Temperatures to Kelvin
We need to convert the temperatures given in Celsius to Kelvin for calculations involving the Coefficient of Performance (COP).

- For the evaporator temperature (\(T_c\)):
\[ T_c = -23^\circ C + 273 = 250 \, K \]

- For the condenser temperature (\(T_h\)):
\[ T_h = 77^\circ C + 273 = 350 \, K \]

**Hint:** Remember that to convert Celsius to Kelvin, you add 273.

### Step 2: Calculate the Coefficient of Performance (COP)
The Coefficient of Performance (COP) of a refrigerator is given by the formula:
\[ COP = \frac{T_c}{T_h - T_c} \]

Substituting the values we found:
\[ COP = \frac{250}{350 - 250} = \frac{250}{100} = 2.5 \]

**Hint:** The COP is a measure of the efficiency of the refrigerator, indicating how much heat is removed per unit of work input.

### Step 3: Calculate the Heat Required to Freeze 1 kg of Water
The heat required to freeze water is given by:
\[ Q_c = m \cdot L_f \]
where:
- \(m = 1 \, kg = 1000 \, g\)
- \(L_f\) (latent heat of fusion of water) is approximately \(80 \, cal/g\) or \(80 \times 4.2 \, J/g\).

Calculating \(Q_c\):
\[ Q_c = 1000 \, g \cdot 80 \, cal/g \cdot 4.2 \, J/cal = 1000 \cdot 80 \cdot 4.2 = 336000 \, J \]

**Hint:** Latent heat is the energy required to change the state of a substance without changing its temperature.

### Step 4: Calculate the Electrical Energy Spent
The work done (electrical energy spent) can be calculated using the relationship:
\[ W = \frac{Q_c}{COP} \]

Substituting the values we have:
\[ W = \frac{336000 \, J}{2.5} = 134400 \, J \]

**Hint:** The work done is the energy input required to remove the heat \(Q_c\) from the system, divided by the efficiency (COP).

### Final Answer
The amount of electrical energy spent in freezing 1 kg of water at \(0^\circ C\) is:
\[ \boxed{134400 \, J} \]