The mess that is ‘signs’ and q = m c ∆T

January 29, 2012

This is messy – really messy – and don’t even get me started on work, internal energy and all that nonsense (which is a catastrophic minefield of bizarre ‘conventions’), but I thought it was time to crystallize a few thoughts on this.

As usual, when studying Thermodynamics with my AP classes, I did a lab that involves a couple of reactions; one endothermic and one exothermic followed by the application of q = m c ∆T. The calculations associated with this lab and these reactions are very messy in two ways. Firstly the problem of deciding what actually constitutes ‘m’ in the equations is tricky. Gases are given off, single replacement reactions lead to the depositing of solids and reactants are in excess – what do we add together to get the correct ‘m’? Secondly, the problem of assigning signs correctly in order to end up with a positive sign for the endothermic reaction and a negative sign for the exothermic reaction is a prickly one. Most texts make much too difficult a song and dance about this IMO, and frankly you often run into the sign convention problem.

The mess that is 'signs' and q = m c ∆TFrankly, the mass ‘issue’ is really a non-issue when it comes to the AP exam (as explained at the end of the document below), but the sign issue has the potential to be much less easy to clean up. This year, for the first time, I decided to create a short document that attempts to clarify things once and for all – I think it worked very well, and it seemed to be the clearest for the kids in many years.

I’m in the habit of keeping EVERYTHING in q = m c ∆T calculations positive (regardless of the energy change actually occurring), and then assigning a sign at the end. For example, in a reaction where there is a temperature change in the water surrounding a chemical reaction, there are two ways to tackle this problem.

Method A: Complicated (generally NOT the way I handle it or teach it)

1. Apply conservation of energy, where if we assume that there are no heat losses, then the heat/energy/enthalpy change in the chemical reaction will be equal and opposite to the heat/energy/enthalpy change in the H2O surrounding the reaction, so;

 qH2O + qreaction = 0

 2. If we re-arrange and substitute we can write;

qH2O = (m) (c) (∆T) = (m) (c) (TempFinal-TempInitial) = – qreaction

(Note the negative sign in front of qreaction)

Consider two situations;

a. If the temperature of the water increases

  • TempFinal > TempInitial, meaning
  • TempFinal – TInitial is a positive number, meaning
  • qH2O is positive (i.e. the water undergoes an endothermic change by absorbing energy), meaning
  • q for the reaction is negative (i.e., the reaction undergoes an exothermic change by releasing energy), meaning
  • the temp of the water goes up (which is where we started).

b. If the temperature of the water decreases

  • TempInitial > TempFinal meaning
  • TempFinal – TInitial is a negative number, meaning
  • qH2O is negative (i.e. the water undergoes an exothermic change by releasing energy), meaning
  • q for the reaction is positive (i.e., the reaction undergoes an endothermic change by absorbing energy), meaning
  • the temp of the water goes down (which is where we started).

Method B: Easier (less ‘scientific’ but usually the way I teach it)

1. Apply qreaction = (m) (c) (∆T), where Delta T is an absolute value so qreaction is always positive initially, then

2. Simply add a positive or negative sign to the q value based upon the knowledge that if,

a. the temp of the water went DOWN then the reaction was endothermic and qreaction must be positive, or

b. the temp of the water went UP then the reaction was exothermic and qreaction must be negative.

Now of course, both of these methods give the correct (same) answer, and I appreciate that the first method is ‘better’ but for ease, you may choose to follow method B.

A quick note about finding the correct mass in q = m c ∆T

You may find it confusing when deciding what to include in the mass for these calculations, but really it’s quite simple. On the AP exam you will likely encounter only one of two situations.

Firstly, one where a mass of water is heated by an EXTERNAL source (as in 1995, 2d). This is easy, since the water is the ONLY thing undergoing the temp. change, so just use its mass.

Secondly, one where chemicals are mixed in solution, and a temp. change occurs in the solution. On the AP exam you will most likely find reactions where there are NO precipitates formed, NO displacement occurs, and NO gas forms, meaning nothing leaves the solution so no mass is ‘lost’, and there are no excess reactants. When this happens, and you are given the specific heat capacity of the solution, use the temp. change of the solution, and use therefore use the mass of the solution, i.e., the sum of the masses of the solute and solvent.


  1. Melanie Dianda

    I couldn’t agree more, I always teach it the easy way. I don’t even bother mentioning the traditional method.

    • c

      Wow, would have been nice if my professor MENTIONED this stuff, because I kept getting positive q values and not understanding that this was the q value for the WATER but not the reaction!

  2. Jenn

    I’m confusing myself a lot with a lab I’m doing. When calculating change in temperature, do you use the highest temperature achieved during the reaction as the “final” value and the initial temperature of the solution, before adding the solute? Are these the two temperature values that give the change in temperature you should use for calculating q?

    • Adrian

      Measure the initial temp of the water, then add the solute and measure the highest temperature reached. ∆T is the difference between those two temperatures. That’s the ∆T in q = m c ∆T.


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