# Heat of mixture of two substances (no state changes)

1. Dec 2, 2012

1. The problem statement, all variables and given/known data

A 100-g sample of Cu(s), initially at 100oC, is added to 140 g of water, initially at 25oC. What is the final temperature of the mixture?

Specific heat
(in J g-1 oC-1)
Cu(s) 0.385
H2O(l ) 4.184

2. Relevant equations

Q = m * ΔT * c

heat lost = heat gained
m1 * ΔT1 * c1 = m2 * ΔT2 * c2

3. The attempt at a solution

100 * (T2 - 100) * 0.385 = 140 * (T2 - 25) * 4.185
38.5* (T2 - 100) = 585.9 * (T2 - 25)
38.5T2 - 3850 = 585.9T2 - 14647.5
T2 = 19.725°C

But the temperature can't be lower than both their starting temperatures, what did I do wrong? The answer is supposed to be 29.6

2. Dec 2, 2012

### Staff: Mentor

Take a look at the expressions you're using for ΔT for the two substances. One should be cooling down while the other is warming up, but you want both to yield positive numbers.

3. Dec 2, 2012

### TSny

You've been snagged by a common sign error for this type of problem. The expression Q = mcΔT represents the heat gained by an object. So, you have to be careful when writing heat lost by an object.

The system as a whole is isolated. So, the total heat gained by the system is zero:

Q1+Q2=0

m1c1ΔT1 + m2c2ΔT2 = 0

Rearrange to

-m1c1ΔT1 = m2c2ΔT2

Note the negative sign on the left (the heat lost term).

So, if you want to set it up as Heat Lost = Heat Gained, then you'll need to handle the sign of the heat lost.

[ASIDE: Did we just set a record for the number of posts within 1 minute???]

Last edited: Dec 2, 2012
4. Dec 2, 2012

100 - t2

5. Dec 2, 2012

### Dick

Heat lost and heat gained should both be positive. You know T2 is between 100 and 25. You have a sign problem. Change (T2-100) to (100-T2).