Water of volume $$2$$ L in a closed container is heated with a coil of $$1$$ kW. While water is heated, the container loses energy at a rate of $$160$$ $$J/s$$. In how much time the temperature of water rise from $$27$$°C to $$77$$°C? (Specific$$ heat of water is $$4.2$$ $$kJ/kg$$ and that of the container is $$negligible).

Question

Infinity Learn · Accepted Answer

Mass of water is ρ$$=mvm=$$ρ$$v=103$$×$$2$$×$$10$$−$$3=2$$  $$kg(since$$ $$1000$$ L $$=$$ $$1m3)Now$$, using Δ$$Q=ms$$Δ$$T=p$$−p'×$$t=2$$×$$4.2$$×$$10377$$−$$27=1000$$−$$160t$$⇒$$t=2$$×$$4.2$$×$$103$$×$$50840=50060s=8$$ min  $$20$$ s.

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time the temperature of water rise from 27°C to 77°C? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

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