Question

What mass of steam at 125 degrees C must be added to 2.50 kg of ice at -28.0 degrees C to yield liquid water at 90.0 degrees F? Show all your work - equations, units, conversions, etc.

Answer 1

m = mass of steam

T = final temperature = 90 deg F = 32.22 degree celcius

Heat lost by steam = Heat gained by ice

m c_steam (125 - 100) + m L_steam + m c_water (100 - 32.22) = m_ice c_ice ( 0 - (-28)) + m_ice L_ice + m_ice c_water (32.22 - 0)

c_steam = specific heat of steam = 2009 J/kgC

c_water = specific heat of water = 4186  J/kgC

c_ice = specific heat of ice = 2093 J/kgC

L_ice = latent heat of fusion from ice to water = 334 x 10³ J/kg

L_steam = latent heat of fusion from steam to water = 2230 x 10³ J/kg

m (2009)(125 - 100) + m (2230 x 10³) + m (4186)(100 - 32.22) = (2.5) (2093) ( 0 - (-28)) + (2.5) (334 x 10³) + (2.5) (4186)(32.22 - 0)

m = 0.514 kg