Temperature of 100 gm water is changed from `0^@C " to " 3^@C` . In this process , heat supplied to water will be (specified heat of water `1 cal g^(-1) .^@C^(-1))`

Options

A. Equal to 300 cal
B. Greater than 300 cal
C. Less than 300 cal
D. Data is insufficient

Answer

To solve the problem of calculating the heat supplied to 100 grams of water when its temperature is changed from \(0^\circ C\) to \(3^\circ C\), we can use the formula for heat transfer:

\[ Q = m \cdot s \cdot \Delta T \]

Where:
- \(Q\) = heat supplied (in calories)
- \(m\) = mass of the substance (in grams)
- \(s\) = specific heat capacity (in cal/g°C)
- \(\Delta T\) = change in temperature (in °C)

### Step 1: Identify the values
- Mass of water, \(m = 100 \, \text{g}\)
- Specific heat capacity of water, \(s = 1 \, \text{cal/g°C}\)
- Initial temperature, \(T_i = 0^\circ C\)
- Final temperature, \(T_f = 3^\circ C\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_f - T_i = 3^\circ C - 0^\circ C = 3^\circ C \]

### Step 3: Substitute the values into the heat transfer formula
\[ Q = m \cdot s \cdot \Delta T \]
\[ Q = 100 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 3^\circ C \]

### Step 4: Perform the calculation
\[ Q = 100 \cdot 1 \cdot 3 = 300 \, \text{cal} \]

### Conclusion
The heat supplied to the water is \(Q = 300 \, \text{cal}\). However, since the specific heat capacity may vary slightly with temperature, the actual heat supplied can be considered to be slightly less than 300 calories.