`75 gm` of copper is heated to increase its temperature by `10^@ C`. If the same quantity of heat is given to 'm' gm of water, to have same rise in temperature is (specific heat of copper `= 420 J//Kg-^@ C`).

Options

A. 7.5 gm
B. 5 gm
C. 10 gm
D. 25 gm

Explanation

To solve the problem, we need to find the mass \( m \) of water that would absorb the same amount of heat as \( 75 \, \text{g} \) of copper when both are heated to achieve a temperature rise of \( 10^\circ C \).

### Step-by-step Solution:

1. **Identify the given values:**
- Mass of copper, \( m_c = 75 \, \text{g} \)
- Temperature rise, \( \Delta T = 10^\circ C \)
- Specific heat of copper, \( s_c = 420 \, \text{J/(kg} \cdot ^\circ C) \)

2. **Convert the specific heat of copper to calories:**
- \( 1 \, \text{cal} = 4.184 \, \text{J} \)
- Therefore, \( s_c = \frac{420 \, \text{J/(kg} \cdot ^\circ C)}{4.184 \, \text{J/cal}} = 100 \, \text{cal/(kg} \cdot ^\circ C) \)
- Since \( 1 \, \text{kg} = 1000 \, \text{g} \), we convert it to:
\[ s_c = \frac{100 \, \text{cal}}{1000 \, \text{g}} = 0.1 \, \text{cal/(g} \cdot ^\circ C) \]

3. **Calculate the heat absorbed by copper:**
- The formula for heat absorbed is:
\[ Q = m \cdot s \cdot \Delta T \]
- For copper:
\[ Q_c = m_c \cdot s_c \cdot \Delta T = 75 \, \text{g} \cdot 0.1 \, \text{cal/(g} \cdot ^\circ C) \cdot 10^\circ C \]
- Calculate \( Q_c \):
\[ Q_c = 75 \cdot 0.1 \cdot 10 = 75 \, \text{cal} \]

4. **Set up the equation for water:**
- Let the mass of water be \( m_w \).
- The specific heat of water, \( s_w = 1 \, \text{cal/(g} \cdot ^\circ C) \).
- The heat absorbed by water is:
\[ Q_w = m_w \cdot s_w \cdot \Delta T \]
- Since we want the same heat absorbed:
\[ Q_c = Q_w \]
- Thus:
\[ 75 \, \text{cal} = m_w \cdot 1 \, \text{cal/(g} \cdot ^\circ C) \cdot 10^\circ C \]

5. **Solve for \( m_w \):**
\[ 75 = m_w \cdot 10 \]
\[ m_w = \frac{75}{10} = 7.5 \, \text{g} \]

### Final Answer:
The mass \( m \) of water required to achieve the same temperature rise is \( 7.5 \, \text{g} \).