Question

10 g of ice at −25°C is heated by a burner which is supplying heat energy at a rate of 250 J/s. Calculate the time, in which the water formed from ice attains a temperature of 90°C. Specific heat capacity of ice is 2.1 Jg⁻¹°C⁻¹ and specific latent heat of ice is 336 Jg⁻¹.

Answer 1

Heat gained by ice to attain the temperature of 0°C = mcθ_R

= 10 × 2.1 × 25

= 525 J

Heat gained by ice to form water at 0°C = mL_ice

= 10 × 336

= 3360 J

Heat gained by water till 90°C = mcθ_R

= 10 × 4.2 × 90

= 3780 J

∴ Total heat gained = (525 + 3360 + 3780) J = 7665 J

Also, rate of supply of heat = 250 J/s

∴ Time required \[=\frac{7665}{250}=\mathrm{3 0 . 6 6 ~ s}\]