Calorimetry And Heat Exchange
A 200 g copper block at 100 °C is dropped into 300 g of water at 20 °C inside an insulated cup. Given copper's specific heat is 0.39 J g−1K−1 and water's is 4.18 J g−1K−1, find the equilibrium temperature.
Select the correct option:
Solution
24.4 °C
When two bodies are mixed in an insulated container, the principle of calorimetry requires that heat lost by the hotter body equals heat gained by the cooler one, since no energy escapes. Let T be the common final temperature. Heat lost by copper is mccc(100−T) and heat gained by water is mwcw(T−20). Setting them equal: 200(0.39)(100−T)=300(4.18)(T−20), which is 78(100−T)=1254(T−20). Expanding gives 7800−78T=1254T−25080, so 32880=1332T and T≈24.4°C. The value 30 °C ignores the large heat capacity of water. The value 60 °C wrongly takes the simple average of the two temperatures. The value 21.5 °C overweights the water and underestimates copper's contribution. This balance assumes the cup is perfectly insulated and that no heat is absorbed by the container itself, so all the energy lost by copper is accounted for by the water. As a plausibility check, water has both more mass and a far higher specific heat than copper, so the equilibrium temperature should lie close to water's initial 20 °C, and 24.4 °C is appropriately near it rather than near the midpoint of the two starting temperatures.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- thermodynamics
- Topic
- calorimetry and heat exchange
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
24.4 °C
When two bodies are mixed in an insulated container, the principle of calorimetry requires that heat lost by the hotter body equals heat gained by the cooler one, since no energy escapes. Let T be the common final temperature. Heat lost by copper is mccc(100−T) and heat gained by water is mwcw(T−20). Setting them equal: 200(0.39)(100−T)=300(4.18)(T−20), which is 78(100−T)=1254(T−20). Expanding gives 7800−78T=1254T−25080, so 32880=1332T and T≈24.4°C. The value 30 °C ignores the large heat capacity of water. The value 60 °C wrongly takes the simple average of the two temperatures. The value 21.5 °C overweights the water and underestimates copper's contribution. This balance assumes the cup is perfectly insulated and that no heat is absorbed by the container itself, so all the energy lost by copper is accounted for by the water. As a plausibility check, water has both more mass and a far higher specific heat than copper, so the equilibrium temperature should lie close to water's initial 20 °C, and 24.4 °C is appropriately near it rather than near the midpoint of the two starting temperatures.
This medium difficulty physics question is from the chapter thermodynamics, covering the topic of calorimetry and heat exchange. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse thermodynamics questions on RankGuru.