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JEE Advanced 2016 · Paper 1, Physics question 3

A water cooler of storage capacity 120 litres can cool water at a constant rate of PP watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30 °C and the entire stored 120 litres of water is initially cooled to 10 °C. The entire system is thermally insulated. The minimum value of PP (in watts) for which the device can be operated for 3 hours is

(Specific heat of water is 4.2 kJ kg−1 K−14.2\ \mathrm{kJ\,kg^{-1}\,K^{-1}} and the density of water is 1000 kg m−31000\ \mathrm{kg\,m^{-3}})

Schematic of a closed circulation loop: a water tank labelled Cooler on the left connected by two pipes to a box labelled Device on the right containing a coiled heat-exchanger; the upper pipe carries Hot water from the device back to the cooler (arrow pointing left), the lower pipe carries Cold water from the cooler to the device (arrow pointing right).
Choose one option
Answer and worked solution

Official answer: B

Heat to remove in 3 hours. 3000×3×3600=32.4×1063000\times3\times3600=32.4\times10^6 J.

What the stored water can absorb. 120 kg warming from 10 °C to the limit of 30 °C:
120×4200×20=10.08×106 J.120\times4200\times20=10.08\times10^6\text{ J}.

The cooler supplies the rest.
P×10800=32.4×106−10.08×106  ⟹  P≈2067 W.P\times10800=32.4\times10^6-10.08\times10^6\implies P\approx2067\text{ W}.

(B)

Topics:Calorimetry and specific heat
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