Energy efficiency for one cup of a hot drink

Bill Davy

Can't get a life till you have finished this research! We can put numbers on these things to get a better idea of efficiency.

Specific heat capacity of water = 4.2 J cm3 K This means it takes 4.2 J to raise the temperature of 1 cm3 (= 1 ml) of water by 1 degree K (= 1 degree C).

So let's say your cold water comes out of the tap at 10 degrees C (after all, you 'let it run till it's cold', right?). A cup of tea is typically 180 ml (at least that's how much water cup a soups (cups a soup?) say to add). So we need a total of

4.2 J cm3 K * 180 cm3 * 90 K = 68040 J

let's say 68kJ.

For the microwave oven, the power (in watts) will be clearly indicated - let's suppose it's a 700W oven (modern microwaves are usually 650-850 W I believe, mostly at the top end for newer ones). In 2m20s, such an oven will output

700W * 140s = 98 kJ

So the microwave outputs 98kJ in order to impart 68kJ to the water - this gives us an efficiency figure of 68-98 ~= 68 plus 1%

For the kettle, the power rating will probably not be so visible - you might have to look the manual, or it *might* be indicated somewhere on the kettle itself. I recently bought one of the new wave of dirt-cheap supermarket kettles, which was advertised as having a 3kW element. I ran some tests on it (yes really), and found it boiled a cup of water in 40s.

3 kW * 40s = 120 kJ

So the efficiency here is 68-120 ~= 57%.

So this hypothetical 700W microwave is roughtly 20% more efficient than my kettle.

This is key to a useful experiment, yes :)

-- Larry Lard Replies to group please