[osake]

A 500g ball of copper at 75ºC is cooled by dropping it into 300g of water at 20ºC. Assuming no heat is lost, what will be the equilibrium temperature?

For both the ball and water the difference in temperature will be the same right? 75-x=20+x, so x=27.5, so the equilibrium temp would be 47.5?

[nicatre]

The mass plays a role, you have more of the warmer substance.

[osake]

(Original post by nicatre)
The mass plays a role, you have more of the warmer substance.

Sorry I don't get your answer can you explain it in more detail?

[Hazza616]

If the weight and temps are all you are given, then i suppose it would be (Mass of Copper * Temp of Copper)+(Mass of water * Temp of water) All over the combined mass.

[Naarim]

(Original post by osake)
Sorry I don't get your answer can you explain it in more detail?

I think he's saying that the mass of the copper ball should be equated for. For example, It'd cool down quicker if it was 100g instead of 500g.

[osake]

(Original post by Naarim)
I think he's saying that the mass of the copper ball should be equated for. For example, It'd cool down quicker if it was 100g instead of 500g.

ah, well see i know that it should be equated for, i tried it yesterday, spent like 30 mins trying and didn't get the answer. Would usually not give up that fast but i cant afford to spend anymore time trying to answer the same question when the physic exams is tomorrow.

[osake]

bump

[Hazza616]

Whats the answer?

[osake]

(Original post by Hazza616)
Whats the answer?

says the answer is 27ºC

[osake]

bump

[Xarren]

At equilibrium Qw = Qc

as Q = mc(t1-t2)

m(copper)c(copper)(t1 - T)=m(water)c(water)(T-t2)

Quite clearly you require the specific heat capacity. Without it this question is impossible.