Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading about the subject of heat in a basic physics book. If I am not mistaken the formula to work out how much energy is required to increase the temperature of water is

e = M * t * shc


  • e is energy in Joules
  • M is mass in kg
  • t is temperature to increase by in °C
  • shc is specific heat capacity in J/kg°C

If I need to solve how much the temperature has varied I rearrange the equation like this...

t = e / M / shc

And this is where I get stuck.

  • e = 7.2 * 108 J
  • M = 105 kg
  • shc = 4.23 J/kg°C

What is the resulting unit of measurement of the following?

e / M = ?
M / shc = ?
share|cite|improve this question
up vote 0 down vote accepted

Just write it out like this:

$t = \frac{e}{\frac{m}{shc}}= \frac{J}{\frac{kg}{\frac{J}{kg\cdot{}^\circ{}C}}} = \frac{J}{J \cdot kg \cdot \frac{1}{kg\cdot^{\circ}C}} = \frac{1}{\frac{1}{^{\circ}C}} = ^{\circ}{\rm{}C}$

share|cite|improve this answer
This was the explanation I found easiest to understand, thanks! – Peter Morris May 22 '13 at 17:13

When multiplying or dividing units, all you need to do is put the units in the numerator or denominator (wherever they appeared) of the answer. So:

$$[e/M]={J\over kg}$$ $$[M/shc]={kg\over{J\over kg^oC}}={kg^2\,^{\circ}\rm C\over J}$$

But this is not the correct way of analyzing your units. You have
$t = e / M / shc = e / (M * shc)$

The units of this are: $$[t]={J\over kg {J\over kg\,^{\circ}\rm C}}=\,^{\circ}\rm C$$

share|cite|improve this answer
Wow, Jim. Apparently your answer was so good that two different people decided to post different halves of your answer afterwards. – Jack Dozer May 21 '13 at 15:40
@Dan Indeed. But everyone is entitled to answer in their own way, so let's not have inflammatory comments like that anymore. Okay? – Jim May 21 '13 at 15:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.