How many calories of heat will be required to convert 1 g of ice at `0^(@)C` into steam at `100^(@)C `

Options

A. 720 cal
B. 640 cal
C. 540 cal
D. 180 cal

Answer

To find out how many calories of heat are required to convert 1 g of ice at 0°C into steam at 100°C, we need to calculate the total heat required in three steps:

1. **Melting the Ice**: We need to convert ice at 0°C to water at 0°C. This process requires the latent heat of fusion.
- The latent heat of fusion of ice is 80 calories per gram.
- For 1 g of ice:
\[ Q_1 = m \cdot L_f = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

2. **Heating the Water**: Next, we need to raise the temperature of the water from 0°C to 100°C. This requires the specific heat of water.
- The specific heat of water is 1 calorie per gram per degree Celsius.
- The temperature change (ΔT) is from 0°C to 100°C, which is 100°C.
- For 1 g of water:
\[ Q_2 = m \cdot c \cdot \Delta T = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 100 \, \text{cal} \]

3. **Vaporizing the Water**: Finally, we need to convert the water at 100°C to steam at 100°C. This process requires the latent heat of vaporization.
- The latent heat of vaporization of water is 540 calories per gram.
- For 1 g of water:
\[ Q_3 = m \cdot L_v = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

Now, we can sum up all the heat quantities to find the total heat required (Q):
\[ Q = Q_1 + Q_2 + Q_3 = 80 \, \text{cal} + 100 \, \text{cal} + 540 \, \text{cal} = 720 \, \text{cal} \]

Thus, the total heat required to convert 1 g of ice at 0°C into steam at 100°C is **720 calories**.

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