10 gm of ice at 0°C is kept in a calorimeter of water equivalent 10 gm. How much heat should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat)

Options

A. 6200 cal
B. 7200 cal
C. 13600 cal
D. 8200 cal

Answer

To solve the problem step by step, we need to calculate the total heat required to convert 10 grams of ice at 0°C into steam at 100°C. This involves three main processes: melting the ice, heating the water to 100°C, and then vaporizing the water.

### Step 1: Calculate the heat required to melt the ice
The heat required to melt ice (H1) can be calculated using the formula:
\[ H_1 = m \times L_f \]
where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion for ice)

Calculating \( H_1 \):
\[ H_1 = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 2: Calculate the heat required to raise the temperature of water to 100°C
After melting, we have 10 grams of water at 0°C. The heat required to raise the temperature of this water (H2) is given by:
\[ H_2 = m \times s \times \Delta T \]
where:
- \( m = 10 \, \text{g} \) (mass of water)
- \( s = 1 \, \text{cal/g°C} \) (specific heat capacity of water)
- \( \Delta T = 100°C - 0°C = 100°C \) (temperature change)

Calculating \( H_2 \):
\[ H_2 = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times 100°C = 1000 \, \text{cal} \]

### Step 3: Calculate the heat required to vaporize the water
The heat required to vaporize the water (H3) is calculated using the formula:
\[ H_3 = m \times L_v \]
where:
- \( m = 10 \, \text{g} \) (mass of water)
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization for water)

Calculating \( H_3 \):
\[ H_3 = 10 \, \text{g} \times 540 \, \text{cal/g} = 5400 \, \text{cal} \]

### Step 4: Calculate the total heat required
Now, we can find the total heat (H_total) required by adding all three heats:
\[ H_{\text{total}} = H_1 + H_2 + H_3 \]
\[ H_{\text{total}} = 800 \, \text{cal} + 1000 \, \text{cal} + 5400 \, \text{cal} = 7200 \, \text{cal} \]

### Final Answer
The total heat required to evaporate the water formed from 10 grams of ice at 0°C is **7200 calories**.

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