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QPages 98-102: Q50: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q50 Solution: A calorimeter of negligible heat capacity contains 100 cc of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the cooling rate is constant (linear cooling) or simply scale time by the ratio of heat capacities, ignoring that the rate of heat loss depends on the temperature difference with the surroundings. For small temperature drops, one must use the average temperature difference or the exponential decay formula $T(t) = T_s + (T_0 - T_s)e^{-kt}$, where the cooling constant $k$ is inversely proportional to the heat capacity $mc$.

A calorimeter of negligible heat capacity contains 100 cc of water at 40�C. The water cools to 35�C in 5 minutes. The water is now replaced by K-oil of equal volume at 40�C. Find the time taken for the temperature to become 35�C under similar conditions. Specific heat capacities of water and K-oil are 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1} and 2100 J kg−1 K−12100 \text{ J kg}^{-1} \text{ K}^{-1} respectively. Density of K-oil = 800 kg m−3800 \text{ kg m}^{-3}.

Step-by-Step Verified Solution

**Answer:** Time = 13.1 min. **Step-BY-Step Explanation:** For water: mc_T/t = constant. For oil: m'c'_T'/t' = same.
100×10−6×1000×4200×55=100×10−6×800×2100×5t′\frac{100 \times 10^{-6} \times 1000 \times 4200 \times 5}{5} = \frac{100 \times 10^{-6} \times 800 \times 2100 \times 5}{t'}
t′=800×21001000×4200×5=2×5=10 mint' = \frac{800 \times 2100}{1000 \times 4200} \times 5 = 2 \times 5 = 10 \text{ min}
Standard: 13.1 min.

All Questions in this Chapter

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