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  4. A water cooler of storage capacity 120 liters can cool water at a constant rate of P watts. In a closed circular 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 <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/63af40657775db32e46cc541c44c79e9-.png /> (Specific heat of water is 4.2 kJ kg -1 K -1 and the density of water is 1000 kg m -3 )

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Q. A water cooler of storage capacity $120$ liters can cool water at a constant rate of $P$ watts. In a closed circular 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
image
(Specific heat of water is $4.2\, kJ kg ^{-1} K ^{-1}$ and the density of water is $1000\, kg m ^{-3}$ )

JEE AdvancedJEE Advanced 2016

Solution:

Rate of heat extracted by water $=\left(\frac{ ms \Delta T }{\Delta t }\right)$
Power by cooler $+\frac{ ms \Delta T }{\Delta t }= H$
$P +\frac{ ms \Delta T }{\Delta t }= H$
For $P _{\min }$ in $\Delta t =3 hrs$
$P = H -\frac{ ms \Delta T }{\Delta t }$
$ P =3 \times 10^{3}-\frac{ P \times 120 \times 10^{-3} \times 4.2 \times 10^{3} \times(30-10)}{3 \times 60 \times 60} $
$ P =2067$ Watt