# The equation of raising the temperature of the water using the electric heater

I've been confused of the following equation

The electric heater exists which can generate $$P[cal/s]$$.

$$t:=$$time which is required to raise the temperature of the water of $$m[kg]$$ from $$a[°C]$$ to $$b[°C]$$.

The $$x$$ percent of a quantity of heat which is generated from the heater will be absorbed into the water.

The below equation is true.

$$P*t*(\frac{x}{100})=m*(b-a)$$

I've googled to find the equation which is as same as above one,however I couldn't.

I've thought that it is weird that the above equation doesn't contain a variable of a specific heat capacity.

Can anyone tell me what I've missing or the site(s) which describes of it?

In a time $$t$$ your heater generates an heat $$W$$ given by $$W=Pt$$ a fraction $$x$$ of this [I am assuming $$x$$ is already divided by 100 so it is a number between 0 and 1] i. e. $$w=xW$$ is adsorbed by water, that is, the heat actually adsorbed by water in a time $$t$$ is

$$w=xW=xPt$$

On the other hand, the heat capacity of water $$c$$ is 4200 J/Kg/C i.e. for $$1kg$$ of water you need 4200 J to raise it by one degree.

So if you start at the initial temperature $$T_0=T(t=0)$$ and you have a mass $$m$$ of water and you want to raise it to $$T(t)=T(t=t)$$ you are going to need a total energy (heat) of $$w$$ $$w=mc \left( \;T(t)-T_0\;\right)$$

equating the two equations you get

$$xPt = mc\left( \;T(t)-T_0\;\right)$$ This is very close to your result, but you had "incorporated" the heat capacity in your definition of $$t$$ as the time needed to heat up your mass of water, so your "time" was actually a normalised time containing information about the heat capacity.

Summing up, the temperature of your water bath, assuming there is no loss, no heat circulation etc. is going to be given by

$$T(t) = {xP\over mc}t+T_0$$