Search any question & find its solution
Question:
Answered & Verified by Expert
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:
2148 Upvotes
Verified Answer
Given, $\Delta Q=100 \mathrm{~W}=100 \mathrm{~J} / \mathrm{s}$
Useful work done, $\Delta W=75 \mathrm{Js}^{-1}$
Now, $\Delta U=\Delta Q-\Delta W=100-75=25 \mathrm{~J} / \mathrm{s}$.
Useful work done, $\Delta W=75 \mathrm{Js}^{-1}$
Now, $\Delta U=\Delta Q-\Delta W=100-75=25 \mathrm{~J} / \mathrm{s}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.