Equation used to find the efficiency of a kettle

[physicsnerd26]

So, I'm doing this coursework right now, and I know that this equation is what's needed to get to the efficiency but I have only a little idea why.

I'm given this formula:
https://s.yimg.com/hd/answers/i/669504ce234c41af99c351e0c28eea10_A.png?a=answers&mr=0&x=1424239496&s=8a1c1557912955a6446f8f8e90ff7a57

It's Efficiency = (mass (volume of water) x water capacity x change in temperature)/ (voltage x current x time taken)

Is there any theory connected to this? Or other equations that led to this equation?

 

[SteamKing]

Perhaps you should look up the definition of 'efficiency'.

Also, don't post duplicate threads in different forums. That's a violation of the rules of PF.

 

[physicsnerd26]

Perhaps you should look up the definition of 'efficiency'.

Also, don't post duplicate threads in different forums. That's a violation of the rules of PF.

Already have, but am I right to assume that:

mass x water capacity x change in temperature = energy output
and the voltage x current x time taken = energy input?

 

[SteamKing]

Already have, but am I right to assume that:

mass x water capacity x change in temperature = energy output
and the voltage x current x time taken = energy input?

That's correct.

 

[DEvens]

I think C has to be the heat capacity of water, not the "water capacity" whatever that may be.

The heat capacity of water will be in units of energy per change in temperature per unit mass. For example, Joules per degree C per kilogram.