Heat gained and lost by aluminium and water in thermal equilibrium

[hkor]

Homework Statement

A physicist pours 100g of water at 100'C into a 20g aluminium cup which already contains 50g of water at 20'C. The final equilibrium temperature of the water is 71.847.
C(water)=4186J/kgC
C(aluminium)=900J/kgC

1. the amount of heat lost by the hot water
2. the amount of heat gained by the aluminium
3. the amount of heat gained by the cold water

Homework Equations

Q=mc(dt)

The Attempt at a Solution

I have attempted to answer them but really have no clue if the answers I am getting are right or if I am even using the correct approach because we weren't given answers or solutions

1: Q=(100e-3)(4186)(71.847-100)=-11784.8
2. I am really unsure about how to do this. Is it relevant that we don't know the initial temperature of the aluminium or can we assume it is in equilibrium with the water inside?
3. Q=(50e-3)(4186)((71.847-20)= 10851.57

Any help would be great, and/or if you wouldn't mind checking my answers. Thanks :)

 

[BruceW]

yep, looks good. I get the same answers as you do. And for question 2), you have the right idea. You assume that the aluminium is initially in equilibrium with the water inside. So you've got the initial temperature, and using a similar reasoning can give you the final temperature.