Question

Calculate the amount of ice at 0°C, which should be added to 250 g of water at 45°C, such that resulting temperature of the mixture is 5°C.

[Specific latent heat of fusion of ice = 336 J g⁻¹]

Answer 1

Assume the specific heat capacity of water = 4.2 J g⁻¹°C⁻¹

Let the mass of ice added be m g.

Heat lost by water = mcΔT

= 250 × 4.2 × (45 − 5)

= 42000 J

Heat gained by ice = Heat used to melt ice + Heat used to warm melted water from 0°C to 5°C

= m × 336 + m × 4.2 × 5

= 336m + 21m

= 357 m J

By the principle of the method of mixtures,

Heat gained = Heat lost

357 m = 42000

m = $$\frac{42000}{357}$$

m = 117.65 g

Therefore, add about 117.6 g of ice at 0 °C.