Q.
Water of volume 2 litre in a container is heated with a coil of 1kW at 27∘C. The lid of the container is open and energy dissipates at rate of 160J/s. In how much time temperature will rise from 27∘C to 77∘C ? [Given specific heat of water is 4.2kJ/kg ]
Heat gained by the water =( Heat supplied by the coil)
- (Heat dissipated to environment) ⇒mcΔθ=PCoil t−PLosst ⇒2×4.2×103×(77−27)=1000t−160t ⇒t=8404.2×105=500sec=8min20sec