Hello, Chris,
To answewr this question, we need to calciulate the energies involved for
1) warming the ice from -50.0C to 0C,
2) melting the ice at 0C, and
3) warming the liquid from 0 to 90C
Since the temperature only gets up to 90.0C, we don't need the heat of vaporization nor the specific heat of steam, although I'll admit they are tempting.
We need to careful with the units. The specific heats come in moles, grams, Joules, and kiloJoules. Very easy to make a mistake, unless we are careful. The answer is requested in kJ, so we'll convert to that unit whenever we get the opportunity. We are given 1 mole of H2O, which is 18 grams. The units we are given include both moles and mass (grams), so we can use either 1 mole or 18 grams, when needed.
In all cases, we calculate the heat involved with the relationship q = cp*m*(DT), where q is the energy, cp is the specific heat, m is the mass (or moles), and DT is the temperature change (Delta T).
1) Warm the ice from -50.0C to 0C
q = (2.09 J/gK)*(18 g)(-50oK - 0oK) = -1881 J
The negative indicates that 1881 Joules are absorbed by the system. Converet this to kJ by dividing by 1000: -1.881 kJ.
2) Melt the ice at 0oC
There is no temperature change for this phase change from solid to liquid. The specific heat of fusion is 6.01 kJ/mole. We have 1 mole, so this will be -6.01 kJ absorbed by the ice in becoming water, all at 0oC.
3) Raise the temperature of water from 0oC to 90oC.
The specific heat of liquid water is 4.18 J/gK. To raise the temperature of 18 grams of water:
q = (4.18 J/gK)*(18 g)*(90-0oC) = -6772 J, or -6.77 kJ
The final step is to add the energies for each step. I get a sum of -14.66, or -14.7 kJ to raise the temperature of 1 mole of ice from -50 to 90oC.
Bob
