JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Heat & Thermodynamics

An electric appliance supplies

6000 J / min

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 \times 1 0^{3} J

(A) $4.1 \times 1 0^{1} s$
(B) $2.4 \times 1 0^{3} s$
(C) $2.5 \times 1 0^{1} s$
(D) $2.5 \times 1 0^{2} s$ ✓ Correct

Answer: (D)

Solution

The appliance supplies heat at

\dot{Q} = \frac{6000}{60} = 100

W while the system delivers work at

\dot{W} = 90

W. By the first law the internal energy grows at

\frac{d U}{d t} = \dot{Q} - \dot{W} = 100 - 90 = 10 W

so an increase of

2.5 \times 1 0^{3}

J needs

t = \frac{2.5 \times 1 0 ^{3}}{10} = 2.5 \times 1 0^{2} s

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option

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