A refrigeratior converts `50` gram of water at `15^(@)C` intoice at `0^(@)C` in one hour. Calculate the quantity of heat removed per minute. Take specific heat of water `=1` cal `g^(-1) .^(@)C^(-1)` and latent heat of ice `=80` cal `g^(-1)`

Explanation

To calculate the quantity of heat removed by the refrigerator when converting 50 grams of water at 15°C into ice at 0°C, we need to consider two processes: cooling the water from 15°C to 0°C and then freezing the water at 0°C into ice.

### Step 1: Calculate the heat removed while cooling the water from 15°C to 0°C.

The formula to calculate the heat removed when cooling is given by:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \( Q_1 \) = heat removed (calories)
- \( m \) = mass of water (grams)
- \( c \) = specific heat of water (cal/g°C)
- \( \Delta T \) = change in temperature (°C)

Given:
- \( m = 50 \) g
- \( c = 1 \) cal/g°C
- \( \Delta T = 15°C - 0°C = 15°C \)

Substituting the values:
\[ Q_1 = 50 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 15 \, \text{°C} \]
\[ Q_1 = 50 \cdot 15 \]
\[ Q_1 = 750 \, \text{cal} \]

### Step 2: Calculate the heat removed while freezing the water at 0°C into ice.

The formula to calculate the heat removed during the phase change (freezing) is given by:
\[ Q_2 = m \cdot L \]
where:
- \( Q_2 \) = heat removed (calories)
- \( L \) = latent heat of fusion (cal/g)

Given:
- \( L = 80 \) cal/g

Substituting the values:
\[ Q_2 = 50 \, \text{g} \cdot 80 \, \text{cal/g} \]
\[ Q_2 = 4000 \, \text{cal} \]

### Step 3: Calculate the total heat removed.

The total heat removed, \( Q \), is the sum of the heat removed during cooling and the heat removed during freezing:
\[ Q = Q_1 + Q_2 \]
\[ Q = 750 \, \text{cal} + 4000 \, \text{cal} \]
\[ Q = 4750 \, \text{cal} \]

### Step 4: Calculate the quantity of heat removed per minute.

Since the total heat removed is done in one hour (60 minutes), we can find the heat removed per minute:
\[ \text{Heat removed per minute} = \frac{Q}{60} \]
\[ \text{Heat removed per minute} = \frac{4750 \, \text{cal}}{60} \]
\[ \text{Heat removed per minute} \approx 79.17 \, \text{cal/min} \]

### Final Answer:
The quantity of heat removed per minute is approximately **79.17 calories**.

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