A water cooler of storage capacity 120 liters can cool water at a constant rate of P 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 liters of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is:
(Specific heat of water is 4.2 kJ kg –1 K –1 and the density of water is 1000 kg m –3 )
Text Solution
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Heat generated in device in 3 hours = 3 × 3600 × 3 × 10 3 = 324 × 10 5 J
Heat used to heat water = ms Δθ = 120 × 1 × 4.2 × 10 3 × 20 J
Heat absorbed by coollant = Pt = 324 × 10 5 – 120 × 1 × 4.2 × 10 3 × 20 J
Pt = (325 – 100.8) × 10 5 J
P = 
= 2067W
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