Two 50-g ice cubes are dropped into 200 g of water in a glass. If the water was initially at a temperature of 25 oC, and if the ice came directly from a freezer operating at a temperature of - 15 oC, what will be the final temperature of the drink?

Final Temperature of the drink = - 15 °C

Explanation:

Heat lost by the water = heat gained by the cubes ... Equation 1

Heat lost by water = c₁m₁ (T₁-T₃) ... Equation 2

Heat gained by the cubes = Latent heat of fusion of the cubes + Heat capacity of the cubes.

⇒ Heat gained by the cubes = 2 (lm₂) + 2{c₂m₂ (T₂-T₃) } ... equation 3

Substituting equation 2 and 3 into equation 1

c₁m₁ (T₁-T₃) = 2 (lm₂) + 2{c₂m₂ (T₃-T₂) } ... Equation 4

Making T₃ The subject of formula in equation 4 above,

T₃ = (c₁m₁T₁ - 2lm₂ + 2c₂m₂T₂) / (c₁m₁ + 2c₂m₂) ... Equation 5

Where c₁ = Specific heat capacity of water, c₂ = specific heat capacity of ice, m₁ = mass of water, m₂ = mass of ice, l = specific latent heat of fusion of ice, T₁ = initial temperature of water, T₂ = initial temperature of ice, T₃ = final temperature of the drink.

Constant: l = 336000 J/kg, c₁ = 4200 J/kg. K, c₂ = 2100 J/kg. K.

Given: m₁ = 200 g = (200/100) kg = 0.2 kg, m₂ = 50 g = (50/1000) kg = 0.05 kg, T₁ = 25 °C = 298 K, T₂ = - 15 °C = 258 K

Substituting these values into equation 5,

T₃ = (4200*0.2*298 + 2*336000*0.05 + 2*2100*0.05*258) / (4200*0.2 + 2*0.05*2100)

T₃ = (250320 - 33600 + 54180) / (840 + 210)

T₃ = 270900/1050

T₃ = 258 K = - 15 °C

Therefore The Final Temperature of the drink = - 15 °C