An electrically heated coil is immersed in a calorimeter containing 360 g of water at 10oC. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and coil is 40 g. The temperature of water after 10 min is

Energy supplied by the heater to the system in 10 minQ1=P×t=90J/s×10×60s=54000J=540004.2cal=12857calNow if θ is the final temperature of the system, energy absorbed by it to change its temperature from 10oC to θoC isQ2=(msΔT)water +(msΔT)coil + calorimeter =360×1×(θ−10)+40(θ−10)=400(θ−10)According to problem Q1=Q2 So 12857=400(θ−10) or θ=42.14∘C