Question:
Water of volume 2 liters in a container is heated with a coil of 1 kW at 27°C. The lid is open, and energy dissipates at a rate of 160 J/s. In how much time will the temperature rise from 27°C to 77°C?
(Given specific heat of water = 4.2 kJ/kgK)
Water of volume 2 liters in a container is heated with a coil of 1 kW at 27°C. The lid is open, and energy dissipates at a rate of 160 J/s. In how much time will the temperature rise from 27°C to 77°C?
(Given specific heat of water = 4.2 kJ/kgK)
Show Hint
Always subtract heat losses from input power.
Updated On: Mar 23, 2026
- 8 min 20 s
- 6 min 2 s
- 7 min
- 14 min
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The Correct Option is A
Solution and Explanation
Step 1: Mass of water = 2kg, Δ T = 50^∘C.
Step 2: Heat required: Q = mcΔ T = 2 × 4200 × 50 = 4.2×10⁵J
Step 3: Effective power: P = 1000 - 160 = 840W
Step 4: Time: t = (Q)/(P) = (4.2×10⁵)/(840) = 500s
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