Certain amount of heat is given to 100 g of copper to increase its temperature by `21^@C`. If same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat capacity of copper ` = 400 J kg^(-1) K^(-1)` and that for water ` = 4200 J kg^(-1) K^(-1))`

Options

A. `4^(@)C`
B. `5.25^(@)c`
C. `8^(@)C`
D. `6^(@)C`

Explanation

To solve the problem, we will use the formula for heat transfer, which is given by:

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

where:
- \( Q \) is the heat supplied,
- \( m \) is the mass,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

### Step 1: Calculate the heat supplied to copper
Given:
- Mass of copper, \( m_{Cu} = 100 \, \text{g} = 0.1 \, \text{kg} \) (conversion from grams to kilograms)
- Specific heat capacity of copper, \( c_{Cu} = 400 \, \text{J/kg/K} \)
- Change in temperature for copper, \( \Delta T_{Cu} = 21 \, \text{°C} \)

Using the formula:
\[ Q = m_{Cu} \cdot c_{Cu} \cdot \Delta T_{Cu} \]
Substituting the values:
\[ Q = 0.1 \, \text{kg} \cdot 400 \, \text{J/kg/K} \cdot 21 \, \text{K} \]
\[ Q = 0.1 \cdot 400 \cdot 21 = 840 \, \text{J} \]

### Step 2: Use the same amount of heat for water
Now, we will use the same amount of heat \( Q \) for the water.

Given:
- Mass of water, \( m_{water} = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Specific heat capacity of water, \( c_{water} = 4200 \, \text{J/kg/K} \)

Using the heat formula for water:
\[ Q = m_{water} \cdot c_{water} \cdot \Delta T_{water} \]
Substituting the known values:
\[ 840 \, \text{J} = 0.05 \, \text{kg} \cdot 4200 \, \text{J/kg/K} \cdot \Delta T_{water} \]

### Step 3: Solve for the change in temperature for water
Rearranging the equation to find \( \Delta T_{water} \):
\[ \Delta T_{water} = \frac{840 \, \text{J}}{0.05 \, \text{kg} \cdot 4200 \, \text{J/kg/K}} \]
Calculating the denominator:
\[ 0.05 \cdot 4200 = 210 \, \text{J/K} \]
Now substituting back:
\[ \Delta T_{water} = \frac{840}{210} = 4 \, \text{°C} \]

### Final Answer
The rise in temperature of the water is \( 4 \, \text{°C} \).

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