Question

A 7-kg cube of copper (c_Cu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.5 kg of water (c_water = 4186 J/kg-K) with an initial temperature of 293 K.

1)

What is the final temperature of the water-and-cube system?

K

2)

If the temperature of the copper was instead 1350 K, it would cause the water to boil. How much liquid water (latent heat of vaporization = 2.26

Answer 1

Q1 348.5 K

Question 2
First start by finding the amount of energy needed to raise the water from 293 to 373 K
373 - 293 = 80 K
80 K * (5.2 kg * 4186 J/kg-K) = 1741376 J needed
Now calculate the amount of energy required to lower the copper from 1350 K to 373 K
1350 - 373 = 977 K
977 K * (7.8 kg * 386 J/kg-K) = 2941551.6 J
Subtract to find energy NOT used to raise water to boiling point
2941551.6 - 1741376 = 1200175.6 J
^^^This value is the amount of energy used to vaporize water. latent heat is 2.26*10^6 J/kg, meaning 2260000 J is required to vaporize 1 kg of water. Divide the energy by the latent heat to find the amount of kilograms of water vaporized.
1200175.6 J / 2260000 = 0.531 kg vaporized
5.2 - 0.531 = 4.669 kg water remains

Question 3
Find amount of energy expended by the copper
750 K - 373 K = 377 K
377 K * 3010.8 J/kg-K = 1135071.6
For the water, find the temperature change and multiply it by the specific heat
(373-293)*4186 = 334880
^^^ This value is the amount of energy required to raise 1 kg of water by (373-293) K. Now we need the amount of energy required to vaporize 1 kg of water, which we already know to be 2260000 J

So I'm having a hard time explaining this part, but I basically just made an algebraic equation based on the information above, m being the variable for the mass of the water.

1135071.6 - 334880m - 2260000m = 0
m = 0.437 kg