Question

Calculate the amount of heat in kJ that is required to heat 20.0 g of ice from -25 ºC to 90 ºC, and sketch a heating curve for the process. The specific heat of ice is 2.11 J/( g. ºC); water 4.18 J/( g. ºC) and the ΔH_fus for water is 6.01 kJ/mol

Answer 1

The "cold water" in this case is going to do three things:

a) as a solid, warm up from -25 to zero
b) all 20 g will melt
c) as a liquid, the temperature goes up 90 Deg Celcius

The warmer water must provide all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder (q_lost = q_gain), we have this:

heat to warm ice 25 degrees + heat to melt ice + heat to warm cold water to 90 Deg celcius

Here are the numbers:

                                        [(20)(25)(2.11)] + [(6010)(25.0/18.0)] + [(25) (90) (4.184)]

                                        1055 + 8347.22 + 9414

                                        = 18816.22 Joules or 18.816 Kilo Joules

18.816 Kilo Joules that is required to heat 20.0 g of ice from -25 ºC to 90 ºC