[Foobarz]

Ok so this question states this:

A 30 g sample of water at 280 K is mixed with 50 g of water at 330 K. Calculate final temperature:

So one way is to use q = -q, which is 30g*s(T_f-280K) = -50g*s(T_f-330K) where s is the specific heat of water (but it doesn't matter). The answer is 311 K. HOWEVER, if I do it using an "average" or

( 30.0 g * 280 K + 50.0 g * 330 K ) / 80.0 g

The answer is still 311 K. BUT HOW??? Does the unit of gram-Kelvin (g * K ) make any sense? Is mass times temperature = energy (since specific heat is measured in J/g*K)?

Answers are kindly appreciated

[Borek]

q=-q and average are mathematically equivalent - the latter can be derived from the former.

Solve q=-q for T_f and you will see (specific heat cancels out)

No idea where you see gK units - in the nominator? Look at the whole equation, not at parts.

[Foobarz]

Yeah you find the g*K units in the numerator, since when you simplify all the numbers you get (g*K) / g = K, so thats why the answer is in K. I know the specific heats cancel out. so in other words

30.0g (4.184 J/g*K)*280K + 50.0g (4.184 J/g*K)*330K = 1
                                   80.0g (4.184 J/g*K)* y

where y is the final temperature

For the left hand side to equal 1, the numerator and denominator must be equal. I think that this way of solving the question is trying to say that g*K is the absolute energy of the two waters combined and then divide the energy (g*K) by the mass (g) to get temperature (K). But I thought it was impossible to find the absolute energy? Or is the 280K and 330K in the numerator mean a change in temperature of +280K and +330K from a reference point of 0K? Can you do that?