Question

Calculate the amount of ice, which is sufficient to cool 150 g of water, contained in a vessel of mass 50 g (S.H.C. = 0.4 Jg⁻¹°C⁻¹) at 30°C, such that the final temperature of the mixture is 5°C.

[Specific latent heat of ice is 336 Jg⁻¹]

Answer 1

Substance	Mass	S.H.C./S.L.H.	Initial Temp.	Final Temp. = 5°C
Ice	?(x)	336 Jg−1	0°C	θR = (5 − 0) = 5°C
Water	150 g	4.2 Jg−1°C−1	30°C	θF = (30 − 5) = 25°C
Vessel	50 g	0.4 Jg−1°C−1	30°C

Heat gained by ice to form water at 0°C = mL_ice = x × 336

Heat gained by water formed from ice = mcθ_R

= x × 4.2 × 5

= 21 x

∴ Total heat gained = 336 x + 21 x = 357 x

Heat lost by water at 30°C = mcθ_F

= 150 × 4.2 × 25

= 15750 J

Heat lost by the vessel at 30°C = mcθ_F

= 50 × 0.4 × 25

= 500 J

Total heat lost = (15750 + 500) = 16250 J

Heat gained = Heat lost

357 x = 16250

\[\text {x (Mass of ice)} =\frac{16250}{357}\]

∴ x = 45.52 g