Question

How much heat is needed to convert 889 g of ice at -10°C to steam at 136°C? (The specific heats of ice, steam, and liquid are 2.03 J/g·°C, 1.99 J/g·°C, and 4.18 J/g·°C, respectively. The heat of fusion is 6.02 kJ/mol and heat of vaporization is 40.79 kJ/mol.) kJ

Answer 1

Total energy is equal to sum of 5 processes

ie

-10 ^oC ice is heated to 0^oC Ice

0^oC ice is melted to 0^oC water

0^oC water is heated to 100 ^oC water

100 ^oC water is vaporised to 100 ^oC steam

100 ^oC to 136 ^oC steam

energy required for the conversion of -10 ^oC ice is heated to 0^oC Ice = mass x specific heat x change in temperature = 889 x 2.03 x 10 = 18046.7 J

energy required for the conversion of 0 ^oC ice is heated to 0^oC water = number of moles x heat of fusion

number of moles of water in 889 g = 889/18 =49.39

Note: molar mass of water is 18 g/mol

energy required = 49.39x 6020 = 297327.8 J

(Note: unit of heat of fusion is converted to J/mole)

energy required for the conversion of 0 ^oC water is heated to 100^oC water = mass x specific heat x change in temperature = 889 x 4.18 x 100 = 371602 J

energy required for the conversion of 100 ^oC water is heated to 100^oC steam = number of moles x heat of vaporization = 49.39 x 40790 = 2014618.1 J

(Note: unit of heat of vaporization is converted to J/mole)

energy required for the conversion of 100 ^oC steam is heated to 136^oC steam = mass x specific heat x change in temperature = 889 x 1.99 x36 = 63687.96 J

total energy = 63687.96 J + 2014618.1 J + 371602 J + 297327.8 J + 18046.7 J = 2765282.56 J

= 2765.28256 kJ