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)

• A. 8 min 20 s
• B. 6 min 2 s
• C. 7 min
• D. 14 min

Hint:

Always subtract heat losses from input power.

Answer 1

The Correct Option is A

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