Heat provided to water will be responsible for increase in temperature. Therefore, we can write
η×P×t=M×s×ΔT
Now, η=10070=0.7 and energy=power×time.
Therefore,
t=0.7×20002×4200×(60∘−10∘)s=300s
JEE Main 2023 — Physics Thermodynamics
A water heater of power 2000W is used to heat water. The specific heat capacity of water is 4200Jkg−1K−1. The efficiency of heater is 70. Time required to heat 2kg of water from 10∘C to 60∘C is _____ s.
(Assume that the specific heat capacity of water remains constant over the temperature range of the water).
Held on 31 Jan 2023 · Verified 6 Jul 2026.
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Work through every JEE Main Thermodynamics PYQ, year by year.