[MJF]

How much heat (in kJ) is needed to convert 866 g of ice at −10°C to steam at 126°C? (The specific heats of ice and steam are 2.03 J/g · °C and 1.99 J/g · °C, respectively.)

The correct answer was 2.67E3 kJ. But here is how I did it based on an example that I looked at:

Step 1: ms(delta t)
(866g) (2.03 J/g · °C) (100°C−10°C) = 158 kJ

Step 2: 866g * 1 mol H2O / 18.02 g H2O * 40.79 kJ/mol / 1 mol H2O = 1960 kJ

Step 3: (886g) (1.99 J/g · °C) (126°C-100°C) = 448 kJ

158 kJ + 1960 kJ + 448 kJ = 2566 kJ

What did I do wrong?

[curiouscat]

Step 1: ms(delta t)
(866g) (2.03 J/g · °C) (100°C−10°C) = 158 kJ

What's this?

[DrCMS]

You need to break this up into 5 different steps.

• Heating ice from -10°C to 0°C (specific heat capacity of ice)
• Melting the ice (latent heat of fusion)
• Heating water from 0°C to 100°C (specific heat capacity of water)
• Boiling the water (latent heat of vapourisation)
• Heating steam from 100°C to 126°C (specific heat capacity of steam)