A 1.0 kW heater supplies energy to a liquid of mass 0.50 kg. The temperature of the liquid changes by 80 K in a time of 200 s. The specific heat capacity of the liquid is 4.0 kJ kg–1K–1. What is the average power lost by the liquid?

I applied Q=mct but I am not still getting the power lost.

1 Answer
Jacob T.
Apr 25, 2018

#P_"loss"=0.20color(white)(l)"kW"#

Explanation:

Start by finding the energy lost over the period of #200color(white)(l)"seconds"#:

#W_"input"=P_"input"*t=1.0*200=200color(white)(l)"kJ"#
#Q_"absorbed"=c*m*Delta*T=4.0*0.50*80=160color(white)(l)"kJ"#

The liquid is going to absorb all the work done as thermal energies if there's no energy loss. The increase in temperature shall equal to #(W_"input")/(c*m)=100color(white)(l)"K"#
However, due to heat transfer, the actual gain in temperature isn't as high. The liquid ended up absorbing only part of the energy; the rest was lost. Therefore:

#W_"lost"= W_"input"-Q_"absorbed"=200-160=40color(white)(l)"kJ"#

Average power equals to work over time, therefore
#barP_"lost"=(W_"lost")/(t)=40/200=0.20color(white)(l)"kW"#

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