How much heat is added to 10.0kg of ice at -20.0C to change it to steam at 120.0C?

Here is a similar question and answer, just change a few numbers. In the SI system, stick with kg. Question: What is the energy needed to heat 1 kg of ice at –20ºC to steam at 150ºC Ans: specific heat of water is 4.186 kJ/kgC specific heat of ice is 2.06 kJ/kgC specific heat of steam is 2.1 kJ/kgK heat of fusion of ice is 334 kJ/kg heat of vaporization of water is 2256 kJ/kg 5 parts to this problem, add them up to get the total to warm ice to 0C E1 = 2.06 kJ/kgC x 1 kg x 20C to melt ice E2 = 334 kJ/kg x 1 kg to warm water from 0C to 100C E3 = 4.186 kJ/kgC x 1 kg x 100C to boil water E4 = 2256 kJ/kg x 1 kg to heat steam from 100C to 150C E5 = 2.1 kJ/kgC x 1 kg x (150–100)C E = E1 + E2 + E3 + E4 + E5

Hi m(kg) not (g) or (gr) Q = mc∆ϴ and Q = mL(f) or Q = mL(c) ice -20 -- To --> ice 0 -- To --> water 0 -- To --> water 120 -- To --> steam 120 Ice to ice and water to water , steam to steam : Q = mc∆ϴ Ice to water : Q = mL(c) water to steam : Q = mL(f) we need "c" and L(f) , L(c)

Hi m(kg) not (g) or (gr) Q = mc∆ϴ and Q = mL(f) or Q = mL(c) ice -20 -- To --> ice 0 -- To --> water 0 -- To --> water 120 -- To --> steam 120 Ice to ice and water to water , steam to steam : Q = mc∆ϴ Ice to water : Q = mL(c) water to steam : Q = mL(f) we need "c" and L(f) , L(c)