If I want to calculate the efficiency of a microwave oven, I have the input Energy (= power (watts) x time (s)) and the output which I measured using temperature difference (q=mct). So do I divide output/input x 100 or do I find the difference between them (=work) then divide it by the input energy?

  • $\begingroup$ Make sure your units balance on both sides... $\endgroup$
    – Solar Mike
    Mar 24, 2019 at 17:25
  • $\begingroup$ yes they do input energy: microwave is 650W and I'm heating for 20 seconds, so 650W x 20s = 13000J output energy: experimental value of 7698.56J $\endgroup$
    – Shahad
    Mar 24, 2019 at 17:28
  • $\begingroup$ so to calc effieicny do I just divide the 2 then x 100? or do I need to find the difference between them then divide by the input energy? $\endgroup$
    – Shahad
    Mar 24, 2019 at 17:30

1 Answer 1


The equation for efficiency in thermodynamics is as follows:

$$\eta = W_{out}/W_{in}$$

and to get this as a percentage you just multiply $\eta \;\mathrm{x}100=\%$

See the link below! https://en.wikipedia.org/wiki/Thermal_ef

  • $\begingroup$ it says its work out / Q in, not work in $\endgroup$
    – Shahad
    Mar 24, 2019 at 18:36
  • $\begingroup$ so in my case... if u read the question, I calculated the energy in using the microwaves power and I calculated the energy in using q=mct... so what do I use???/////// $\endgroup$
    – Shahad
    Mar 24, 2019 at 18:37
  • 2
    $\begingroup$ You should be able to apply the answer to your situation.... How about "efficiency = Useful Output / Total input" $\endgroup$
    – Solar Mike
    Mar 24, 2019 at 19:04
  • $\begingroup$ I'm agree with @SolarMike, his/her definition is more accurate, the useful output is the temperature of the food, but the inside and the food container (plates or...) also the magnetic losses in transformer is not negligible, and you didn't take the effect of reactive power into account, since the microwave can be indicative. I'm sure the apparent power gives you more accurate results, in case of capacitive microwave the active power is a good estimation. $\endgroup$
    – user14407
    Mar 24, 2019 at 20:08
  • $\begingroup$ @Shahad The link actually says efficiency = benefit / cost.... before you get down to Q or W... $\endgroup$
    – Solar Mike
    Mar 24, 2019 at 20:28

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