JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Heat & Thermodynamics

A water cooler of storage capacity 120 litres 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 litres 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}

)

(A)1600
(B)2067✓ Correct
(C)2533
(D)3933

Answer: (B)

Solution

The net heat absorbed by the water over 3 hours equals the heat generated by the device minus the heat removed by the cooler:

(P_{heater} - P_{cooler}) \times t = m s Δ T

where

m = 120

kg,

s = 4.2 \times 1 0^{3}

J/(kg·K),

Δ T = 20

K, and

t = 3 \times 3600

(3000 - P) \times 3 \times 3600 = 120 \times 4.2 \times 1 0^{3} \times 20

(3000 - P) \times 10800 = 10.08 \times 1 0^{6}

3000 - P = 933.33

P = 2066.67 \approx 2067

watts

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option

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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Heat & Thermodynamics

Solution

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