Heat and Thermodynamics 1 Question 10
10. 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 $30^{\circ} C$ and the entire stored 120 litres of water is initially cooled to $10^{\circ} C$. The entire system is thermally insulated. The minimum value of $P$ (in watts) for which the device can be operated for 3 hours is
(2016 Adv.)
(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
(c) 2533
(d) 3933
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Solution:
- Heat generated in device in $3 h$
$$ =\text { time } \times \text { power }=3 \times 3600 \times 3 \times 10^{3}=324 \times 10^{5} J $$
Heat used to heat water
$$ =m s \Delta \theta=120 \times 1 \times 4.2 \times 10^{3} \times 20 J $$
Heat absorbed by coolant
$$ \begin{aligned} & =P t=324 \times 10^{5}-120 \times 1 \times 4.2 \times 10^{3} \times 20 J \\ P t & =(325-100.8) \times 10^{5} J \\ P & =\frac{223.2 \times 10^{8}}{3600}=2067 W \end{aligned} $$
