When 60 calories of heat are supplied to 15 g of water, the rise in temperature is

Options

A. `75^@C`
B. `900^@C`
C. `4^@C`
D. `0.25^@C`

Explanation

To solve the problem of finding the rise in temperature when 60 calories of heat are supplied to 15 g of water, we can use the formula:

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

Where:
- \( Q \) is the heat supplied (in calories),
- \( m \) is the mass of the substance (in grams),
- \( s \) is the specific heat capacity (in calories per gram per degree Celsius),
- \( \Delta T \) is the change in temperature (in degrees Celsius).

### Step-by-Step Solution:

1. **Identify the given values:**
- Heat supplied, \( Q = 60 \) calories
- Mass of water, \( m = 15 \) g
- Specific heat of water, \( s = 1 \) calorie/g°C (this is a known constant for water)

2. **Rearrange the formula to solve for \( \Delta T \):**
\[ \Delta T = \frac{Q}{m \cdot s} \]

3. **Substitute the known values into the equation:**
\[ \Delta T = \frac{60 \text{ calories}}{15 \text{ g} \cdot 1 \text{ calorie/g°C}} \]

4. **Calculate the denominator:**
\[ 15 \text{ g} \cdot 1 \text{ calorie/g°C} = 15 \text{ calories/°C} \]

5. **Now perform the division:**
\[ \Delta T = \frac{60 \text{ calories}}{15 \text{ calories/°C}} = 4 \text{ °C} \]

6. **Conclusion:**
The rise in temperature of the water is \( 4 \text{ °C} \).

### Final Answer:
The rise in temperature when 60 calories of heat are supplied to 15 g of water is **4 °C**.