Phase Change Problem!

  • Thread starter DanielT
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  • #1
I am working on a proble with this basic set up:

500g of some material starts at -22'C in a 400g container. 90g of steam (water) starts at 120'C and is mixed with the unknown material. Find the final temp. and amount of material and water in each state.

c(solid) = 0.4 cal/g'c
Melting = 20'c
c(liquid) = 0.9 cal/g'c
Boiling = 80'c
c(vapor) = 0.6 cal/g'c
Hf = 100 cal/g
Hv = 600 cal/g
***Specific Heat of container is 0.2 cal/g'c

Can someone point me in the right direction, OR work it out if you are so inclined!

Answers and Replies

  • #2
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Well its a matter of starting with the sample at 500 g and the container of 400 g, at -22°C and bringing both to 20°C, so delta-T=2°C. Calculate that energy by mass * specific heat of each. That heat comes from the steam, which cools from 120° to some temperature. If above 100°C then it is still superheated steam. If the T is less than 100°C (assuming 1 atm of pressure), then set T=100°C, so delta-T = 20°C, and the remainder of the heat would come from condensation from vapor to liquid.

As vapor cools to liquid, it does so at constant temperature, in the process of condensation. The opposite, vaporization, also occurs at constant temperature. One has to determine the amount of steam (vapor) and liquid. The fraction of steam is called the quality.

As a substance melts from solid to liquid, it does so at constant temperature. Melting is the opposite of freezing or solidification, and heat of fusion is the thermal property.

Where the sample melts at 20°C, assume all heat is absorbed by the unknown, while the container stays at 20°C.