# 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
Apr 25, 2018

${P}_{\text{loss"=0.20color(white)(l)"kW}}$

#### Explanation:

Start by finding the energy lost over the period of $200 \textcolor{w h i t e}{l} \text{seconds}$:

${W}_{\text{input"=P_"input"*t=1.0*200=200color(white)(l)"kJ}}$
${Q}_{\text{absorbed"=c*m*Delta*T=4.0*0.50*80=160color(white)(l)"kJ}}$

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:

${W}_{\text{lost"= W_"input"-Q_"absorbed"=200-160=40color(white)(l)"kJ}}$

Average power equals to work over time, therefore
${\overline{P}}_{\text{lost"=(W_"lost")/(t)=40/200=0.20color(white)(l)"kW}}$