I wanted to find out what percentage of the power supplied by my
microwave ends up in the food as heat. It was tricky because I did
not have a thermometer. I used the following procedure.
Measuring input power:
- Unplugged the refrigerator and checked that no computers or the
stove were running.
- Timed how long it took for the wheel on the power meter to
revolve once.
- Put some chili in a bowl in the microwave, turned it on for three
minutes.
- Timed how long it took for the wheel on the power meter to
revolve once
Measuring output power:
- Emptied a tray of ice into a glass bowl and added
room-temperature water.
- Waited till the water felt as cold as the ice.
- Poured the water through a strainer into a measuring cup. Measured
the volume of water.
- Poured the water back into the bowl. Covered the bowl with
a lid.
- Put the bowl in the microwave and set the timer for 1 or
2 minutes. Waited for the bell to ring.
- Took out the bowl, poured the water through a strainer into the
measuring cup and measured the volume.
Data
Input power
| Trial |
Time with Mic. Off (s) |
Time with Mic. On (s) |
| 1 |
74 |
17 |
| 2 |
77 |
19 |
Output power
| Trial |
Time (s) |
Init Vol of Water (mL) |
Final Vol of Water (mL) |
| 1 |
120 |
500 |
600 |
| 2 |
120 |
125 |
330 |
| 3 |
60 |
330 |
410 |
Calculations
Power meter formula from
Michael Bluejay.
Base Power = ( 3.6 * 7.2 ) / (Time with Mic. Off)
Total Power = ( 3.6 * 7.2 ) / (Time with Mic. On)
Input Power = Total Power - Base Power
This gives 1.17 kW, 1.03 kW. The average is 1.1 kW.
Output Power = ((Final Vol of Water) - (Init Vol of Water))
* (1.0 g/mL) * (333.55 J/g) / (Time)
This gives 0.28 kW, 0.60 kW, 0.445 kW. The average is 0.44 kW
This gives an efficiency of 40%. Unfortunately, the 90% t confidence
interval for the Output Power is (0.17, 0.77) so more trials are
needed.