q(sys) = -q(surr)

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Anne 2L
Posts: 35
Joined: Fri Sep 29, 2017 7:05 am

q(sys) = -q(surr)

Postby Anne 2L » Sat Jan 20, 2018 9:38 pm

I understand why q(sys) + q(surr) = 0.

I was wondering how we can determine if we use

-q(sys) = q(surr)

or

q(sys) = -q(surr)

Michelle Lee 2E
Posts: 64
Joined: Thu Jul 27, 2017 3:01 am

Re: q(sys) = -q(surr)

Postby Michelle Lee 2E » Sat Jan 20, 2018 9:49 pm

I think determining which it is:
-q(sys) = q(surr)
or
q(sys) = -q(surr)
is based on what the question is asking.
If the reaction is exothermic, then it would probably be the former because the system is releasing heat and that heat is positively transferred into the surroundings.
If the reaction is endothermic, then it would probably be the latter because the system is absorbing heat so the surroundings around it would, in turn, lose heat.

If I'm wrong about which kind of reaction coincides with which form of the equation, PLEASE correct me.

Phi Phi Do 2E
Posts: 27
Joined: Fri Sep 29, 2017 7:05 am

Re: q(sys) = -q(surr)

Postby Phi Phi Do 2E » Sat Jan 20, 2018 9:54 pm

The zeroeth Law of Thermodynamics states that heat flows from the hotter object to the cooler object. When it is flowing from one object to another, it flows from the system to the surroundings. When releasing heat, it is an exothermic process which is a negative value and so you would use the q(sys)= -q(surr) because the surroundings are losing heat. Identify the system and the surroundings and so an endothermic process would be -q(sys)=q(surr) because the surroundings are gaining heat.

Sarah Wax 1G
Posts: 29
Joined: Sat Jul 22, 2017 3:00 am

Re: q(sys) = -q(surr)

Postby Sarah Wax 1G » Sun Jan 21, 2018 7:01 pm

Usually, we just use -q(sys) = q(surr).

That way everything remains simple.

If we solve for the heat of system and we get q= -10 kJ, we know that the q of the surroundings is +10 kJ, and therefore the reaction of the system is exothermic. If the opposite signs occur, the reaction of the system is endothermic.


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