An 8.5 g ice cube is placed into 255 g of water. Calculate the temperature change in the water upon the complete melting of the ice. Assume that all of the energy required to melt the ice comes from the water.

Change in temperature of water = 2.67 °C

Explanation:

Heat gained by ice = heat lost by water

Q₁ = Q₂ ... Equation 1

Where Q₁ = Latent heat of ice, Q₂ = heat lost by water

Q₁ = lm₁ ... equation 2

Q₂ = cm₂ΔT ... Equation 3

Substituting equation 2 and 3 into equation 1

lm₁ = cm₂ΔT ... Equation 4

Making ΔT the subject of formula in equation 4

ΔT = lm₁/cm₂ ... Equation 5

Where ΔT = change in temperature of water, l = specific latent heat of ice, m₁ = mass of ice, c = specific heat capacity of water, m₂ = mass of water.

Given: m₁ = 8.5 g = 0.0085 kg, m₂ = 255 g = 0.255 kg.

Constant: l = 336000 J/kg, c = 4200 J/kg. K

Substituting these values into equation 5

ΔT = (336000*0.0085) / (0.255*4200)

ΔT = 2856/1071

ΔT = 2.67 °C

Change in temperature of water = 2.67 °C