Question

100 gm of copper is heated to increase its temperature by $21^\circ C$. If the same amount of heat is given to 50 gm of water, then the rise in its temperature is (Specific heat of Cu = $400~J~kg^{-1}K^{-1}$, Water = $4200~J~kg^{-1}K^{-1}$):

• A. $4^\circ C$
• B. $5.25^\circ C$
• C. $8^\circ C$
• D. $6^\circ C$

Hint:

Ensure units are converted to SI (kg) correctly before calculating.

Answer 1

The Correct Option is A

Step 1: Understand the problem.
Heating 100 g of copper raises its temperature by 21 degrees. The same amount of heat is then given to 50 g of water. We find how much the water heats up.

Step 2: Write the heat formula.
The heat absorbed is $Q = m s \Delta T$, where $m$ is mass, $s$ is specific heat, and $\Delta T$ is the temperature rise.

Step 3: Set the two heats equal.
Since the same heat goes to water as was given to copper, $m_{Cu} s_{Cu} \Delta T_{Cu} = m_{w} s_{w} \Delta T_{w}$.

Step 4: Put in the copper side.
With $m_{Cu} = 0.1$ kg, $s_{Cu} = 400$, $\Delta T_{Cu} = 21$: heat $= 0.1 \times 400 \times 21 = 840$ J.

Step 5: Write the water side.
With $m_{w} = 0.05$ kg and $s_{w} = 4200$: water side $= 0.05 \times 4200 \times \Delta T_{w} = 210 \,\Delta T_{w}$.

Step 6: Solve for the water rise.
Set $840 = 210\, \Delta T_{w}$, so $\Delta T_{w} = 4$ degrees. Water heats less because its specific heat is large. \[ \boxed{4^\circ \text{C}} \]