JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
An electric appliance supplies 6000 J/min 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 × 10³ J ?
- (a) 2.5 × 10² s
- (b) 4.1 × 10¹ s
- (c) 2.4 × 10³ s
- (d) 2.5 × 10¹ s
Answer and solution
Answer (a) 2.5 × 10² s
Solution
Physics sir · 4 steps
Put heat in and work out on the same per-second footing, and their difference fills up the internal energy.
- Same units first. Heat comes in at 6000 J per minute. Divide by 60: that is 100 J s⁻¹ = 100 W.
- First law per second. The system does work at 90 W, so 90 J leaves every second as work. What stays behind raises the internal energy.
(Δ U)/t = Q/t - W/t = 100 - 90 = 10 J s⁻¹
- Find the time. t = (2.5 × 10³ J)/(10 J s⁻¹) = 250 s.
- So. It takes 2.5 × 10² s, which is option (a).
The idea to keep. Write Δ U = Q - W as rates: net power into internal energy = heat supply rate minus power delivered. Convert everything to J/s before subtracting.
Answer from the NTA final answer key. Solution by PrepWiser.