In the equation Q = mc(deltaT), I learned that Q means heat transfer. Is it the heat transfer from the system to the surroundings, i.e. the system loses/absorbs heat or is it from the surroundings to the system? A couple of places on the Internet say that Q is the heat energy of the system... That's different from heat "transfer", isn't it?

Question

if Q is the heat energy of the system, then the equation, deltaH = -Q/n, doesn't make sense to me. Aren't you trying to find the deltaH of the system? So you'd want to use the Q of the system, right? If Qsystem = mc(deltaT) and you put a negative in front of the Q when solving for deltaH, then doesn't that make it the Q of the surroundings rather than the Q of the system (because Qsystem = Qsurroundings ?) so then you'd basically be doing deltaH = Qsurroundings/n. but shouldn't you be doing Qsyatem/n?
I initially thought the negative was in the equation for deltaH because I thought that when you do Q=mc(deltaT) you're getting the Q for the surroundings and you need to make it negative to make it Qsystem to make it work for deltaH. But apparently that's not the case...

I think I sound really unclear, it's hard to describe my confusion but here's an example

A 20.0 g sample of aluminium is cooled and the heat absorbed is equal to 140 J. The specific heat capacity of aluminium is 0.900 J/g•°C. What is the final temperature for this sample if its initial temperature was 20.0°C?
a) 7.8°C
b) 12.3°C
C) 32.2°C
D) 13.5°C
E) this question cannot be answered unless the amount of water used for the cooling process is known.

The answer is b).
my understanding:
Here, Q = -140J, and that's how much energy the aluminium / the system released. m is the mass of the system : 20.0 g. c is the specific heat capacity of the system : 0.900. Initial temperature is the initial temperature of the surroundings but that's equal to the system's temperature? so basically everything in this equation is from the system and Q is the Q of the system.
Anyway, I know how to rearrange the equation and which values to use to get b) as the answer, but for example if this question then asked for enthalpy change, then I'd have to do deltaH = -Q/n, right? And for Q I'd use the value above, -0.140 kJ. But that value is for Qsystem, isn't it? So once a negative goes there, 140 would be the value for Qsurroundings. But deltaH is for the system, so shouldn't you be using Qsystem??

Answer 1

The example problem I gave is the root of all my confusion... Before, I always though when you do Q=mc(deltaT), you're getting the amount of heat absorbed/released by the SURROUNDINGS, and so when you want to get deltaH, you'd have to make Q negative in order to get Qsystem, (because Qsystem = -Qsurroundings), and you want the deltaH of the system, so that's why the equation is deltaH= -Q/n ....

Answer 2

I avoid some of that confusion by other means but let me try to address your questions. I think that most of us mean the system when we talk about Q or heat transfer or anything else in that context EXCEPT when we say specifically, q surroundings, from the surroundings, to the surroundings, etc. With regard to the problem, I don't worry too much about the sign at the BEGINNING; however, you know it is negative. If the system (the Al object) is losing heat it means heat is being emitted (lost) and by definition that is an exothermic reaction and q = -. If I completely ignore the sign, however, and write 140 = 20*0.9*dT and solve for dT, I obtain dT = 140/18 = 7.8. If we know it is losing heat and we know it started at 20, then we know to subtract and get a smaller number than 20. If it asks for delta H, we say it is negative BECAUSE it is losing heat and exothermic is a negative dH. I try not to worry too much about Qsurroundings vs Qsystem; we know if we know Qsystem we can always calculate Qsurroundings. And we always know the sign of Qsystem because it is ALWAYS + if heat is absorbed by the system and - is heat is leaving the system.