# calculating the efficiency of a microwave oven

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?

• Make sure your units balance on both sides... Mar 24 '19 at 17:25
• 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 Mar 24 '19 at 17:28
• 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? Mar 24 '19 at 17:30

## 1 Answer

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

• it says its work out / Q in, not work in Mar 24 '19 at 18:36
• 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???/////// Mar 24 '19 at 18:37
• You should be able to apply the answer to your situation.... How about "efficiency = Useful Output / Total input" Mar 24 '19 at 19:04
• 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. Mar 24 '19 at 20:08
• @Shahad The link actually says efficiency = benefit / cost.... before you get down to Q or W... Mar 24 '19 at 20:28