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 = $4200~J~kg^{-1}K^{-1}$):
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 = $4200~J~kg^{-1}K^{-1}$):
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Ensure units are converted to SI (kg) correctly before calculating.
Updated On: Jun 10, 2026
- $4^\circ C$
- $5.25^\circ C$
- $8^\circ C$
- $6^\circ C$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Heat $Q = ms\Delta T$. Since $Q$ is same, $m_c s_c \Delta T_c = m_w s_w \Delta T_w$.
Step 2: Analysis
$0.1 kg \times 400 \times 21 = 0.05 kg \times 4200 \times \Delta T_w$. $840 = 210 \times \Delta T_w \implies \Delta T_w = 4^\circ C$.
Step 3: Conclusion
The rise in temperature of water is $4^\circ C$.
Final Answer: (A)
Heat $Q = ms\Delta T$. Since $Q$ is same, $m_c s_c \Delta T_c = m_w s_w \Delta T_w$.
Step 2: Analysis
$0.1 kg \times 400 \times 21 = 0.05 kg \times 4200 \times \Delta T_w$. $840 = 210 \times \Delta T_w \implies \Delta T_w = 4^\circ C$.
Step 3: Conclusion
The rise in temperature of water is $4^\circ C$.
Final Answer: (A)
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