The heat of fusion of water is `79.5 ` cal `//` g. This means 79.5 cal of energy are required to

Options

A. raise the temperature of 1g of water by 1K.
B. turn 1 g of water to steam.
C. raise the temperature of 1 g of ice by 1K.
D. melt 1 g of ice.

Answer

To solve the question regarding the heat of fusion of water, we need to understand what the term "heat of fusion" means and how it applies to the conversion of ice to water.

### Step-by-Step Solution:

1. **Understanding Heat of Fusion**:
- The heat of fusion of a substance is the amount of energy required to change it from a solid to a liquid at its melting point without changing its temperature. For water, this value is given as 79.5 calories per gram.

2. **Energy Requirement**:
- This means that to convert 1 gram of ice (solid) at 0°C to 1 gram of water (liquid) at 0°C, you need to supply 79.5 calories of energy.

3. **Physical State Change**:
- The process of changing ice to water is called fusion. During this process, the temperature of the substance does not change; instead, the energy is used to break the bonds between the molecules in the solid state.

4. **Conclusion**:
- Therefore, the statement "79.5 cal of energy are required to melt 1 gram of ice" is correct. It specifically refers to the energy needed to change the state of 1 gram of ice to water without raising the temperature.

### Final Answer:
The heat of fusion of water means that 79.5 calories of energy are required to melt 1 gram of ice into water at 0°C.

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