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JEE Advanced 2016 Paper 1

Section: Physics

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Question : 4 of 54

Marks: +1, -0

A water cooler of storage capacity 120 liters can cool water at 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 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\;\mathrm{kJ}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}

and the density of water is

1000\;\mathrm{kg}\,\mathrm{m}^{-3}

[JEE Adv 2016 P1]

1600
2067
2533
3933

Solution:

3

000 - P = (120 \times 1)(4.2 \times 10^3)\frac{dT}{dt}

\frac{dT}{dt} = \frac{20}{60 \times 60 \times 3}

P = 2067

W

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