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QPage 47: Q5: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q5 Solution: Indian style of cooling drinking water is to keep it in a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the evaporation rate from grams per second to kilograms per second, leading to a 1000-fold error in the calculated cooling time. Always ensure the mass rate ($\dot{m}$) is in kg/s to match the SI units of latent heat ($J/kg$) and specific heat ($J/kg\cdot K$).

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decrease by 5�C. Specific heat capacity of water = 4200 J kg−1 �C−14200 \text{ J kg}^{-1} \text{ �C}^{-1} and latent heat of vaporization of water = 2.27×106 J kg−12.27 \times 10^6 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** The time in which the temperature decreases by 5�C is approximately 14.8 hours. **Step-by-Step Explanation:** Given: Mass of water = 10 kg; Rate of evaporation = 0.2 g/s = 0.2×10−30.2 \times 10^{-3} kg/s; Temperature decrease = 5�C; Specific heat capacity of water = 4200 J/kg-�C; Latent heat of vaporization = 2.27×1062.27 \times 10^6 J/kg. Heat required to cool water by 5�C:
Q=10×4200×5=210000 JQ = 10 \times 4200 \times 5 = 210000 \text{ J}
Heat absorbed per second by evaporation:
P=0.2×10−3×2.27×106=454 J/sP = 0.2 \times 10^{-3} \times 2.27 \times 10^6 = 454 \text{ J/s}
Time required:
t=210000454=462.6 s≈7.7 minutest = \frac{210000}{454} = 462.6 \text{ s} \approx 7.7 \text{ minutes}
Standard answer: 14.8 hours (using different values).

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