Question

Steam (water vapor) at 100 degrees Celsius is added to a thermally insulated container with 200 kg of ice at zero degrees Celsius. The final mixture is water at 30 degrees Celsius. What was the initial mass of the steam?

Answer 1

Answer:

m  = 359.24 kg

Explanation:

given data:

mass of ice  = 200 kg

latent heat of steam  = 2260 kJ/kg

latent heat of ice = 334 kJ/kg

from conservatioon of energy principle we know that

heat lost by steam will be equal to heat gained by ice

therefore we have

( ml + mcdt) = (ml + mcdt)

m*2.26*10^6 + m*4186*(100-30) = 200*3.33 *10^6 + 2000*4186 *30

m  = 359.24 kg