Test Index

Thermodynamics

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Question : 8 of 10
Marks: +1, -0
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:  
Given that
Heat supplied ΔQ=100 W=100 Js−1\Delta Q = 100 \, \mathrm{W} = 100 \, \mathrm{Js}^{-1}
Useful work done ΔW=75 J s−1\Delta W = 75 \, \mathrm{J} \, \mathrm{s}^{-1}
Change internal energy ΔU=?\Delta U = ?
According to first law of thermodynamics, change in internal energy is given by
ΔU=ΔQ−ΔW\Delta U = \Delta Q - \Delta W
ΔU=100−75\Delta U = 100 - 75
=25 J s−1=25 W= 25 \, \mathrm{J} \, \mathrm{s}^{-1} = 25 \, \mathrm{W}
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