Question 41 of 190

An electric appliance supplies $6000J{\min }^{-1}$, heat to the system. If the system delivers a power of $90W$. How long it would take to increase the internal energy by $2.5\times {10}^{3}J$?

A$2.5\times {10}^{1}s$B$2.5\times {10}^{2}s$C$2.4\times {10}^{3}s$D$4.1\times {10}^{1}s$

Show answer and solution▾

Answer: (B) $2.5\times {10}^{2}s$

Recall the formula of the first law of thermodynamics in terms of change in internal energy and work done by the system, $\Delta U=Q-W$, given, $\Delta U=2500J$, $Q=\dfrac{6000}{60}\times t=\left(100t\right)J$ and $W=P\times t=90t$,
$\Rightarrow 2500J=100tJ-90tJ \Rightarrow t=2.5\times {10}^{2}s$.