Question

A 8.6-kg cube of copper (c_Cu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.1 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 × 10⁶ J/kg) will be left after the water stops boiling?

kg

3)

Let's try this again, but this time add just the minimum amount of water needed to lower the temperature of the copper to 373 K. In other words, we start with the cube of copper at 750 K and we only add enough water at 293 K so that it completely evaporates by the time the copper reaches 373 K. Assume the resulting water vapor remaining at 373 K. How much water do we need ?

kg

Answer 1

Question 1

Since its in equilibrium:
-Q_copper = Q_water
-(m_Cu_*cCu_*deltaT) = m_water*_cwater*deltaT
-((8.4)(386)(x-750) = (5.1)*(4186)*(x-293)
-(3242.4x - 2431800) = 21348.6x - 6255132
-3242x + 2431800 = 21348.6x - 6255132
8686932 =24626.6x
x = 352.74K = Final Temp

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 * (5.1 kg * 4186 J/kg-K) = 1707888J needed
Now calculate the amount of energy required to lower the copper from 1350 K to 373 K
1350 - 373 = 977 K
977 K * (8.6 kg * 386 J/kg-K) = 3243249.2 J
Subtract to find energy NOT used to raise water to boiling point
3243249.2  - 1707888 = 1535361.2 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.
1535361.2 J / 2260000 = 0.679 kg vaporized
5.1 - 0.679= 4.421kg 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