Question

A well-insulated styrofoam bucket contains 142 g of ice at 0 °C. If 24 g of steam at 100 °C is injected into the bucket, what is the final equilibrium temperature of the system?(cwater=4186 J/kg.°C, Lf=3.35 × 105 J/kg, Lv=2.26 × 106 J/kg)

Answer 1

Let t be the common temperature attained by the system

The heat absorbed by ice = heat released by steam

Heat change for conversion of ice at 0oC to water at 0 oC + heat change for the conversion of water at 0oC to water at toC = Heat change for the conversion of steam at 100oC to water at 100 oC + heat change for the conversion of water at 100 oC to water at t oC

mL_f + mcdt = m'L_v + m'c'dt'

Where

m = mass of ice = 142 g

m' = mass of steam = 24 g

L_f=heat of fusion of water = 3.35 ×10⁵ J/kg = 3.35x10² J/g = 335 J/g

L_v= heat of vapourization of water = 2.26 × 10⁶ J/kg = 2.26x10³ J/g = 2260 J/g

c = specific heat capacity of water = 4186 J/kgoC = 4.186 J/ goC

dt = t - 0 = t ^oC

dt' = 100-t ^oC

Plug the values we get t = 24.05 oC

Therefore the final equilibrium temperature of the system is 24.05 oC