A well lagged copper calorimeter of mas 120g contains 70g of water and 10g ice both at 0°C. Dry steam at 100°C is passed in until the temperature of the mixture is 40°C. Calculate the mass of the steam condensed. (Specific latent heat of fusion of ice = 3.2 * 10^2) (specific latent heat of vaporization = 2.2 * 10^3) (specific heat capacity of copper = 4.0*10^-1) (specific heat capacity of water = 4.2)

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C - 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C - 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C - 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C - 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g