Posted by **Jena** on Friday, November 17, 2006 at 7:02pm.

the pressure is one atmosphere and determine the heat in joules required to produce 5.03 kg of water vapor at 100.0 °C, starting with (a) 5.03 kg of water at 100.0 °C and (b) 5.03 kg of liquid water at 0.0 °C.

for part (a) I used the equation Q=mL and L was a given value but I tried the same for part (b) but couldn't get the right answer.

For part be, remember the heat to vaporize is at a constant temp of 100C. First you have to h eat the water to that temperature.

Heattotal= heatwater + heat vaporization

= mc(100) + mL

For part b the water must be heated from 0 degrees C to 100 C then added to the heat required to vaporize it at 100 degrees C. You will need to look up the heat of vaporization.

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