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 $$3$$ kW of heat (thermal$$ $$load). The temperature of water fed 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 $$4.2$$ kJ kg−$$1K$$−$$1$$ and the density of water is $$1000$$ kg m−$$3$$ $$)$$

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 $$3$$ kW of heat (thermal$$ $$load). The temperature of water fed 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 $$4.2$$ kJ kg−$$1K$$−$$1$$  and the density of water is $$1000$$ kg m−$$3$$ $$)$$

Infinity Learn · Accepted Answer

Heat generated at the device $$=$$ heat lost at the cooler $$+$$ Heat absorbed by $$water(3kW)$$ (3$$ $$hours) $$=$$ P (3$$ $$hours) $$+$$ ms (30$$ – $$10)⇒$$3$$×$$103$$×$$3$$×$$3600=P$$×$$3$$×$$3600+120$$×$$4200$$×$$20$$​⇒$$P=27$$−$$8.49103=2067W$$

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