A 3kW kettle boils half a litre of water in 70 seconds. What was the temperature of the cold water

Pretty simple. And this is real, i just don't have a thermometer handy!

Please ignore any heat loss from the kettle.

Ok, I've never seen a GCSE student use apostrophes correctly, so you're believed :P

First you want to figure out the energy input (3kW for 70s = 210kJoule).
The mass of water is about 0.5kg (to very good approximation because water is almost exactly 1kg per litre density).
The final temperature will be about 100Celsius, this is about 373Kelvin.

You then need the equation dQ=Cm(dT) where dQ is the change in thermal energy, dT the change in temperature.

dQ = 210kJoule.
C is the heat capacity of water (about 4.2kJoule / kg-Kelvin).

So you now rearrange for dT:

dT = dQ/Cm

Seems to work out to about 0 Celsius. Probably not as low as that since the mass of water will be a little less than 0.5kg in practise.

Yeah, I heard that too - something about since the original methods for measuring calorie content was to see how hot a fixed amount of water got for a fixed amount of "stuff" burning and heating it.

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