thermochemistry

Question

How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?

(The specific heat of ice = 2.10 J/g°C; the specific heat of water = 4.184 J/g°C; the heat of fusion = 333 J/g°C).

Answer 1

This involves 3 steps.

1) heating ice from -23.0 to 0; 2.10 J/g deg X 23.0 deg = 48.3 J/g

2) melting ice at 0; 333 J/g (note the heat of fusion should not have degrees in the units--the temp is constant

3) heating water from 0 to 18.0; 4.184 J/g deg X 18.0 deg = 75.312 J/g

So the total energy is 48.3 + 333 + 75.312 = 456.612 J/g, and multiply by 129 g to get 58902.948 J = 58.9 kJ when rounded off.

You could also multiply by 129 at each step to get J, and then add--the result should be the same.