[pfnm]

108g of water at a temperature of 22.5oC is mixed with 65.1 g of water at an unknown temperature, the final temperature of the resulting mixture is 47.9oC. What was the temperature of the other sample of water?

I'm not sure how to tackle this.

I've thought about it. The second sample of water (65.1g) provides the heat. So the first sample gains heat and the second sample loses heat.

However, for the q=mCΔT equation, I cannot determine mass.

Do I calculate the samples separately? Any advice would be appreciated.

[pfnm]

I tried doing it this way:

ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC.

Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules

Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used

q=mCΔT

11466.58 Joules = 65.1g x 4.18 J / g C x ΔT

11466.58/(65.1gx4.18)=ΔT

ΔT=42.14oC

So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).

Would this be the correct way of approaching the problem?


q=mCΔT

[Borek]

OK

[BluRay]

108g of water at a temperature of 22.5oC is mixed with 65.1 g of water at an unknown temperature, the final temperature of the resulting mixture is 47.9oC. What was the temperature of the other sample of water?

I'm not sure how to tackle this.

I've thought about it. The second sample of water (65.1g) provides the heat. So the first sample gains heat and the second sample loses heat.

However, for the q=mCΔT equation, I cannot determine mass.

Do I calculate the samples separately? Any advice would be appreciated.

Just a little question: why did you post it in a chemistry forum?