how many joules of heat are needed to change 50.0 grams of ice at 150°c to steam at 120°c?

Question

1) how many joules of heat are needed to change 50.0 grams of ice at -15.0°C to steam at 120°C?

2) you have a sample of H20 with a mass 23.0 grams at a temperature of -46.0°C. How many kilojoules of heat energy are necessary to carry out each step? Also, pleade calculate the total amount of energy needed.

a) heat the ice to 0.0°C?

b) melt the ice?

C) heat the water from 0.0°c to 100°c?

d) boil the water?

e) heat the steam from 100°c to 109.0°c?

Answer 1

heat to raise temp of 50 g ice from -15 to 0 deg = (50g)(2.1J/g/deg)(15deg) = 1575 J

heat to melt 50 g ice at 0 deg = 50g x 336 J/g = 16,800 J  -  NOTE: no ∆T during phase change

heat to raise temp of 50 g liquid water from 0 to 100 deg = (50g)(4.2J/g/deg)(100 deg) = 21,000 J

heat to change 50 g liquid to steam at 100 deg = 50 g x 2300 J/g = 115,000 J  - NOTE: not ∆T during phase change

heat to convert 50 steam from 100 to 120 deg = (50g)(1.895 J/g/deg)(20 deg) = 1895 J

Add up all the joules to get total energy needed.  You didn't supply some of the constants, such as specific heat for steam or for ice, etc., so I used commonly used values.

 

Do #2 in a similar fashion.