thermodynamics: ice melting question?

Question

How much heat is required to change 1.0 kg of ice, originally at -20.0°C, into steam at 130.0°C??

Answer 1

Specific heat of each phase times mass, /_\T outside phase transitions. Latent enthalpies of fusion and vaporization each times mass . Add 'em all up.

Answer 2

First, raise the ice to the melting point. Let ice, liquid water, and steam each be called W. Assume that the heat capacity of W(ice) is 0.6cal/g.

1000gW x 20degC x 0.6cal/g-C = 12,000 cal

Next, melt the ice. Assume the heat of fusion of W(ice) is 80cal/g

1000gW x 80cal/1gW = 80,000 cal

Next, raise the water from 0C to 100C. Assume that the heat capacity of water is 1cal/g-C.

1000gW x 100degC x 1cal/g-C = 100,000 cal

Finally, raise the water vapor (steam) from 100C to 130.0C. That will take a rise of 30degC.

I do not know the heat capacity of steam. Let us call it N cal/degC. When you know a value, just plug it in.

1000gW x 30degC x Ncal/g-C = 30,000N cal.

So: You add all the numbers up:
12,000 cal
80,000 cal
100,000 cal
30,000N cal

192,000 cal + 30,000N cal

Answer 3

Use the formula C = m x Q x deltaT

m = mass of water
Q = heat released or absorbed
deltaT = Change in temperature(For water that is)

4.184kJ = 1kg x Q x 110

Q is the variable that is needed for the calculation for heat absorbed or required that is. :) You should be able to solve for Q. Hope this helps. :)

Answer 4

Okay, you need to take 5 things into account:

Specific Heat Capacity of Ice = 2.114 kJ/K kg
Specific Heat Capacity of water = 4.19 kJ/K kg
Specific Heat Capacity of Water vapor = 2.08 kJ/ K kg
Latent Heat of Melting = 5.9 kJ/mol
Latent heat of Vaporisation = 40.7 kJ/mol

OK, lets do some maths:

20 degrees to melting = 20 * 2.114 = 42.28 kJ
100 degrees to boiling = 100 * 4.19 = 419 kJ
30 degrees further heating = 30 * 2.08 = 42.4 kJ
1kg = 55.56 mols of water (1000/18)
55.56 * 5.9 = 327.78 kJ
55.56 * 40.7 = 2261.11 kJ

Add it all up:

3092.57 kJ = 3.092 MJ