Question

40 g of ice at −10°C is heated by a heater of power 250 W, such that water formed from it, attains the temperature equal to the boiling point of water. For how long is the heater switched on?

[S.H.C. of ice = 2 Jg⁻¹°C⁻¹; Sp. latent heat of ice = 340 Jg⁻¹]

Answer 1

Power of heater = 250 W = 250 J/s.

Heat needed to warm ice from −10°C to 0°C:

Q₁ = mc_iceΔT

= 40 × 2 × 10

= 800 J

Heat needed to melt ice at 0°C:

Q₂ = mL

= 40 × 340

= 13600 J

Heat needed to warm water from 0°C to 100°C (boiling point of water):

Q₃ = mc_waterΔT

= 40 × 4.2 × 100

= 16800 J

Total heat needed (Q) = 800 + 13600 + 16800 = 31200 J

Time for which the heater is switched on:

$$t = \frac{Q}{P}$$

$$= \frac{31200}{250}$$

= 124.8 s

Therefore, the heater is switched on for 124.8 s.