how much heat ( in kilojolules) is released for converting ...?

Question

100.0 liters of water steam collected at 127.0 degrees celsius ( under 1 atm) to ice at -20.0 degrees celsius. (Molar mass: O = 16.00 g, H = 1.008 g. The specific heat of ice is 2.03 J/g-degrees celsius, the specific heat of water is 4.184 J/ g- degrees celsius, the specific heat of steam is 1.99 J/g-degrees celisius. the heat of fusion is 6.01 kilojoules for water, and he heat of vaporization is 40.79 kilojoules for water)


please show the steps

Answer 1

Step 1: find the mass of the water (steam) in grams
From the CRC Handbook of Chemistry and Physics, and extrapolating to 127C, the density of water at 127C is 0.9413 gm / ml. There are 100,000 ml of water steam, so there will be 94,130 grams of water.

Step 2. Now, there are 5 regions in which heat will be given up. In each region the RATE at which heat is given up is different.
Region 1 -- going from steam to the point where the water starts to turn to liquid, and losing 27 degrees C of temperature
Region 2 -- going from the vapor phase to the liquid phase, with no change in temperature (remaining at 100 C)
Region 3 -- going from liquid to the point where the water starts turning solid, and losing 100 C of temp.
Region 4 -- going from the liquid phase to the solid phase with no change in temperature (remaining at 0 C)
Region 5 -- going from 0 C to -20 C as a solid.

All of the heat losses in each region must be calculated, then all of them added together to get the total heat.

Region 1 -- steam from 127 C to 100 C
The specific heat of steam is 1.99 J/g C
94,130 g * (127 C - 100 C) * 1.99 J / g C = 5.057 MJ That's 5.057 * 10^6 J

Region 2 -- latent heat of vaporization
40.79 kJ ./ g * 94,130 g = 3.84 GJ thats 3.84 * 10^9 J

Region 3 -- liquid water from 100 C to 0 C
4.184 J / g C * (100 C - 0 C) * 94,130 g = 39.4 MJ

Region 4 and 5... I think you get the idea. Calculate these and then add up all the regions

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