# What would be the final temperature when 100 grams of water at 30C is mixed with 50 grms of water at 60C?

Don’t trust me on this, cause it’s been a while since I’ve done this stuff, but I think it’d be 37.5C because if there were equal amounts of water the temperature would be halfway in between 60C and 30C (45C), but because there’s half as much of the 60C, it’d be closer to 30C, by half of the difference between 30C and 60C. Again, don’t take my word for it, but if I remember right, that’s what it should be. Good luck!

Your equation should be set like this. Q_lost = Q_gained.  Think of it as the 30 degree water gaining the heat that the 60 degree water is losing. SO:

Q=MC(Tf – Ti)

M = Mass of water
C= Specific Heat of Water
Tf = Final temperature of water
Ti = Initial temperature of water

Water has a mass of 1g/ml
Specific heat of water is 1 cal/gram * C

SO

(50)(1.0)(60 – Tf) = (100)(1.0)(Tf – 30)

(50)(60 – Tf) = (100)(Tf – 30)

3000 – 50Tf = 100Tf – 3000

6000 – 50Tf = 100Tf

6000 = 150Tf

Tf = 40  degrees Celsius.