50 g of copper is heated to increase its temperature by 10°C. If the same quantity of heat is given to 10 g of water, the rise in its temperature is (Specific heat of copper = 420 Joule-kg⁻¹°C⁻¹ )

Correct Option: The correct answer is 5°C.

To find the rise in temperature of water, we can use the principle of heat transfer. The quantity of heat energy ($Q$) absorbed or released by a substance of mass $m$ and specific heat capacity $c$ during a temperature change $\Delta T$ is given by the formula:
$Q=m\cdot c\cdot \Delta T$

Step 1: Calculate the heat energy given to the copper
We are given the following values for copper:
Mass of copper ($m_{\text{Cu}}$) = 50 g = $50\cdot {10}^{-3}\text{kg}=0.05\text{kg}$
Rise in temperature ($\Delta T_{\text{Cu}}$) = 10°C
Specific heat of copper ($c_{\text{Cu}}$) = 420 J kg^-1°C^-1

Substitute these values into the heat formula to find the heat energy supplied:
$Q_{\text{Cu}}=m_{\text{Cu}}\cdot c_{\text{Cu}}\cdot \Delta T_{\text{Cu}}$
$Q_{\text{Cu}}=0.05\text{kg}\cdot 420{\text{J kg}}^{-1}{\text{°C}}^{-1}\cdot 10\text{°C}$
$Q_{\text{Cu}}=210\text{J}$

Step 2: Calculate the rise in temperature of the water
The same quantity of heat ($Q=210\text{J}$) is given to the water. We are given the following values for water:
Mass of water ($m_{\text{w}}$) = 10 g = $10\cdot {10}^{-3}\text{kg}=0.01\text{kg}$
Specific heat of water ($c_{\text{w}}$) = 4200 J kg^-1°C^-1

Using the heat formula for water:
$Q_{\text{w}}=m_{\text{w}}\cdot c_{\text{w}}\cdot \Delta T_{\text{w}}$

Substitute the known values to solve for the temperature rise ($\Delta T_{\text{w}}$):
$210=0.01\cdot 4200\cdot \Delta T_{\text{w}}$
$210=42\cdot \Delta T_{\text{w}}$
$\Delta T_{\text{w}}=\frac{210}{42}=5\text{°C}$

Therefore, the rise in temperature of the water is 5°C.