Question

How much heat is required to convert 5.0 kg of ice from a temperature of - 10 0C to water vapor at a temperature of 220 degrees F?

Answer 1

Temperature in ^oC = (F - 32)*5/9

C = (220 - 32)*5/9

= 104.44 ^oC

specific heat capacity of ice = 2040 J/ kg K

specific heat capacity of water = 4186 J/ kg K

Specific heat capacity of steam = 1996 J/kg K

Latent heat of fusion of ice = 334000 J/kg

Latent heat of vaporization of steam = 2260000 J/kg

heat required = mcT for ice + mL_ice + mcT for water + mL_vap + mcT for steam

= 5* (4186*10 + 334000 + 4186*100 + 2260000 + 1996*(104.44 - 100))

= 15316655.6 J

= 15316.65 kJ