A sample of steam with a mass of 0.553 g and at a temperature of 100 C condenses

Question

A sample of steam with a mass of 0.553 g and at a temperature of 100 C condenses into an insulated container holding 4.25 g of water at 5.0 C.

Part A

Assuming that no heat is lost to the surroundings, what will be the final temperature of the mixture?

A sample of steam with a mass of 0.553 g and at a temperature of 100 C condenses into an insulated container holding 4.25 g of water at 5.0 C.

Part A

Assuming that no heat is lost to the surroundings, what will be the final temperature of the mixture?

Explanation / Answer

By using the heat of vaporization of water and the specific heat of water.

(0.522 g) x (2260 J/g) = 1179.72 J lost by the condensing steam.

Supposing all the heat lost by the steam heated the water in the cup then the heat energy change is
q = (1179.72 J) / (4.186 J/g·°C) / (4.25 g) = 66.3°C change
5.0°C + 66.3°C = 71.3°C

So now the problem becomes: Find the final temperature when 0.522 g of water at 100°C is mixed with 4.25 g of water at 71.3°C:

((0.522 g) x (100) + (4.25 g) x (71.3)) / (0.522 g + 4.25 g) = 74.4°C .

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