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11 October, 02:20

A well lagged copper calorimeter of mass 120g contains 70g of water and 10g of ice both at 0 degrees Celsius. Dry steam at 100 degrees Celsius is passed in until the temperature of the mixture Is 40 degrees Celsius. Calculate the mass of steam condensed.

*specific latent heat of fusion of ice=3.2*10²J/g

*specific latent heat of vaporisation of steam=2.2*10³J/g

*specific heat capacity of copper=4.0*10 (raised to power minus one) J/g/k

Specific heat capacity of water=4.2J/g/k

All grams to be changed to kilograms

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  1. 11 October, 02:27
    0
    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
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