Question

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 $3kW$ of heat (thermal load). The temperature of water fed into the device cannot exceed ${30^ \circ }C$ and the entire stored $120$ liters 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(Specific heat of water is $4.2kJk{g^{ - 1}}{K^{ - 1}}$ and the density of water is $1000kg{m^{ - 3}}$ )

\left( A \right)1600 \\\ \left( B \right)2067 \\\ \left( C \right)2533 \\\ \left( D \right)3933 \\\

Answer 1

Hint : In order to solve this question, we are going to first find the heat generated by the device then, the heat that is used in heating water is calculated, then finally the heat that is absorbed by the coolant, then the minimum power that can be generated by operating device for three hours is calculated.
The heat generated in the device in three hours is
$H = P' \times T'$
Heat used in heating water
$H' = ms\Delta \theta$
Heat absorbed by coolant
${H_a} = H - H' = P' \times t - ms\Delta \theta$

Complete Step By Step Answer:
As it is given that the initial temperature of the cooler is
${T_1} = {10^ \circ }C$
And the final temperature of the cooler is
${T_2} = {30^ \circ }C$
And the volume of water is
$V = 120L$
Now the heat generated in the device in three hours is
$H = P' \times T'$
On putting the values in this, we get
$H = 3 \times {10^3} \times 3 \times 60 \times 60 = 324 \times {10^5}J$
The heat that is used in heating water is
H' = ms\Delta \theta = 120 \times 1 \times 4.2 \times {10^3} \times \left( {30 - 10} \right) \\\ \Rightarrow H' = 100.8 \times {10^5}J \\\
The heat absorbed by the coolant is given by
{H_a} = H - H' = P' \times t - ms\Delta \theta \\\ \Rightarrow {H_a} = 324 \times {10^5} - 100.8 \times {10^5} = 223.2 \times {10^5}J \\\
Thus, the minimum power for which the device can be operated for three hours is
$P = \dfrac{{223.2 \times {{10}^5}}}{{60 \times 60}} = 2067watt$
Hence, option $\left( B \right)2067$ is the correct answer.

Note :
As we can see here, that the amount of heat generated is used for two processes, one for heating the water and the rest amount of heat is used for heating the coolant, then finally the minimum power for which the device can operate for three hours is calculated from the heat absorbed by the coolant divided by the time.