Question

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?

Answer 1

Answer:

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