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A calorimeter of negligible heat capacity holds 100 cc of water at 40°C, which cools to 35°C in 300 s. The water is then replaced by an equal volume of K-oil at 40°C. Under the same cooling conditions, how long does the K-oil take to cool from 40°C to 35°C? (Specific heats: water 4200 J/kg-K, K-oil 2100 J/kg-K; density of K-oil = 800 kg/m³.)

1. 120 sec
2. 240 sec
3. 360 sec
4. 480 sec

Correct answer: 120 sec

Solution

For the same loss rate and temperature drop, t ∝ m*c. (m*c)_water = 0.1*4200 = 420; (m*c)_oil = 0.08*2100 = 168; t_oil = 300*(168/420) = 120 s.

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