Specific Heat And Calorimetry
In a calorimetry experiment 200 g of water at 20 degrees Celsius is mixed with 100 g of water at 80 degrees Celsius, and heat exchange with surroundings is neglected.
Select the correct option:
Solution
40 degrees Celsius
The principle of calorimetry in NCERT Class 11, Chapter 11 states that when two bodies at different temperatures are mixed in an isolated system, heat lost by the hotter body equals heat gained by the colder body. Let the final equilibrium temperature be T. Heat gained by the cold water is m1c(T−20)=200c(T−20), and heat lost by the hot water is m2c(80−T)=100c(80−T); the specific heat c cancels. Setting gain equal to loss: 200(T−20)=100(80−T), so 2(T−20)=80−T, giving 2T−40=80−T, hence 3T=120 and T=40 degrees Celsius. The value 50 degrees Celsius is wrong because it takes the simple arithmetic mean, ignoring the unequal masses. The value 60 degrees Celsius is wrong because it weights the calculation toward the hotter, lighter sample by mistake. The value 30 degrees Celsius is wrong because it undershoots by mishandling the mass ratio. In a real calorimetry experiment the calorimeter vessel itself absorbs some heat, so its water equivalent must be added to the cold-side mass; here it is neglected as stated, simplifying the balance. The specific heat cancelling out is a special feature of mixing the same substance, water with water; had different substances been mixed, their differing specific heats would have to be retained in the equation. A plausibility check confirms the result lies between 20 and 80 and, being closer to 20, correctly reflects that the colder water has twice the mass of the hotter water.
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About This Question
- Subject
- physics
- Chapter
- experimental skills
- Topic
- specific heat and calorimetry
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
40 degrees Celsius
The principle of calorimetry in NCERT Class 11, Chapter 11 states that when two bodies at different temperatures are mixed in an isolated system, heat lost by the hotter body equals heat gained by the colder body. Let the final equilibrium temperature be T. Heat gained by the cold water is m1c(T−20)=200c(T−20), and heat lost by the hot water is m2c(80−T)=100c(80−T); the specific heat c cancels. Setting gain equal to loss: 200(T−20)=100(80−T), so 2(T−20)=80−T, giving 2T−40=80−T, hence 3T=120 and T=40 degrees Celsius. The value 50 degrees Celsius is wrong because it takes the simple arithmetic mean, ignoring the unequal masses. The value 60 degrees Celsius is wrong because it weights the calculation toward the hotter, lighter sample by mistake. The value 30 degrees Celsius is wrong because it undershoots by mishandling the mass ratio. In a real calorimetry experiment the calorimeter vessel itself absorbs some heat, so its water equivalent must be added to the cold-side mass; here it is neglected as stated, simplifying the balance. The specific heat cancelling out is a special feature of mixing the same substance, water with water; had different substances been mixed, their differing specific heats would have to be retained in the equation. A plausibility check confirms the result lies between 20 and 80 and, being closer to 20, correctly reflects that the colder water has twice the mass of the hotter water.
This hard difficulty physics question is from the chapter experimental skills, covering the topic of specific heat and calorimetry. It appeared in the 2025 exam.
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