Answer:
Change in temperature of water = 2.67 °C
Step-by-step 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