Question

Water of volume 2 liter in a container is heated with a coil of 1 kw at 27°C. The lid of the container is open and energy dissipates at the rate of 160 (J/S). In how much time temperature will rise from 27° C to 77°C. Specific heat of water is 4.2 {(KJ)/ (Kg)} 

(A) 7 min 

(B) 6 min 2s 

(C) 14 min 

(D) 8 min 20 S

Answer 1

Correct option: (D) 8 min 20 S

Explanation:

Heat gained by water = Heat supplied by coil – Heat dissipated to environment.

∴ m ∙ c ∙ Δθ = (P_coil × t) – (P_loss × t)

Given: Δθ = 77 – 20 = 50°C

C = 4.2 × 10³ J/kg

P_coil = 1 KW = 1000 W

P_loss = 160 W (energy dissipation rate)

m = 2 kg

∴   2 × 4.2 × 10³ × 50 = 1000t – 160t

4.2 × 10⁵ = 840t

 ∴  t = 500 sec = 8 min 20 sec