Q 11-11-129JEE MainJEE Main 2021 (26 Aug, Shift 1)Easy
An electric appliance supplies $6000$ J min$^{-1}$, heat to the system. If the system delivers a power of $90$ W. How long it would take to increase the internal energy by $2.5\times10^3$ J?
Answer: (B) $2.5\times10^2$ s
Heat supplied per second $= \dfrac{6000}{60} = 100$ W; work done by the system $= 90$ W.
Rate of increase of internal energy $= 100 - 90 = 10$ W.
$t = \dfrac{2500}{10} = 250$ s $= 2.5\times10^2$ s.
Solution by Sreeraj P, M.Sc Physics