- #1

FountainDew

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## Homework Statement

A heavy copper pot of mass 2.0 kg (including the copper lid) is at a temperature of 150°C. You pour 0.10 kg of water at 25°C into the pot, then quickly close the lid of the pot so that no steam can escape. Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings.

m(copper) = 2.0 kg

T(copper) = 150°C

c(copper) = 390 J/kg*K

m(water) = 0.10 kg

T(water) = 25°C

c(water) = 4190 J/kg*K

Lv(water) = 2256x10^3 J/kg

## Homework Equations

Heat transfer: Q = mc(delta)T

Heat transfer in Phase Change: Q = m*(Lv)

Conservation of energy: Q(pot) + Q(water) = 0

## The Attempt at a Solution

I calculated the final T to be 106°C for the pot/water system, that was the easy part. The next part says the real final Temp is 100°C because the energy that would have gone into heating the water, actually went into changing its phase from liquid to gas. I don't understand how to do this part at all. I tried calculating a second Q for a (delta)T of 6°C and got a value in Joules. Then I plugged this energy into Q = m*(Lv) and got a mass of 1.1g water vapor which is wrong. So my approach is flawed, why?

**4. The actual solution**

The final temperature is 100°C, and of the 0.10kg of water in the pot, 0.0034kg or 3.4g of the water was changed into vapor.