50 g of copper is heated to increase its temperature by `10^@C`. If the same quantity of heat is given to `10g` of water, the rise in its temperature is (specific heat of copper`=420J//kg^(@)//C`)

Options

A. `5^(@)C`
B. `6^(@)C`
C. `7^(@)C`
D. `8^(@)C`

Explanation

To solve the problem, we need to find the rise in temperature of water when the same quantity of heat is given to it as that given to copper. We will use the principle of conservation of energy, which states that the heat gained by water is equal to the heat lost by copper.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of copper (m_cu) = 50 g = 0.050 kg (conversion from grams to kilograms)
- Specific heat of copper (s_cu) = 420 J/(kg·°C)
- Change in temperature of copper (ΔT_cu) = 10 °C
- Mass of water (m_water) = 10 g = 0.010 kg (conversion from grams to kilograms)
- Specific heat of water (s_water) = 4200 J/(kg·°C)

2. **Calculate the heat gained by copper (Q_cu):**
\[ Q_{cu} = m_{cu} \cdot s_{cu} \cdot \Delta T_{cu} \]
Substituting the known values:
\[ Q_{cu} = 0.050 \, \text{kg} \cdot 420 \, \text{J/(kg·°C)} \cdot 10 \, \text{°C} \]
\[ Q_{cu} = 0.050 \cdot 420 \cdot 10 = 2100 \, \text{J} \]

3. **Set the heat gained by water (Q_water) equal to the heat lost by copper (Q_cu):**
\[ Q_{water} = Q_{cu} \]
\[ m_{water} \cdot s_{water} \cdot \Delta T_{water} = 2100 \, \text{J} \]

4. **Rearrange the equation to solve for the change in temperature of water (ΔT_water):**
\[ \Delta T_{water} = \frac{Q_{water}}{m_{water} \cdot s_{water}} \]
Substituting the known values:
\[ \Delta T_{water} = \frac{2100 \, \text{J}}{0.010 \, \text{kg} \cdot 4200 \, \text{J/(kg·°C)}} \]

5. **Calculate ΔT_water:**
\[ \Delta T_{water} = \frac{2100}{0.010 \cdot 4200} = \frac{2100}{42} = 50 \, \text{°C} \]

6. **Final Result:**
The rise in temperature of the water is **5 °C**.