# Problem with an electricity / thermodynamics assignment

I've been trying to figure this one out for a while on my own, so I'd like to ask for your help if you could offer some.

A heater made out of a wire with a diameter $R = 0.2\text{ mm}$, length $4\pi\text{ m}$ and electrical resistivity of $0.5\times 10^{-6}\ \Omega\;\mathrm{m}$ is connected to a voltage source of $220\text{ V}$, sinked in the water.

Which mass of water will it heat up from $20^{\circ}\mathrm{C}$ to $50^{\circ}\mathrm{C}$ in the time of 10 minutes? (C of water = $4200\ \mathrm{J\;kg}/\mathrm{K}$)

I know I have the electrical properties of the wire and the thermodynamic properties of the water, but I don't know how to proceed from there. We've been studying electricity and I am not really aware how I can connect it with thermodynamics?

-
Hi Phystudent and welcome to Physics Stack Exchange! This is actually not a homework help site, it's for conceptual questions about physics, so your question in its original form was not appropriate for the site. I tried to fix it up and make it more focused on the underlying concepts and more generally useful. I'm not sure how successful I was, but I think as it is now, it can perhaps stay open. If you have any problems with the changes I made, you're free to make further edits, or you can add a comment. –  David Z Oct 21 '11 at 4:22
The three answers below boil down to the same method. –  Fingolfin Oct 21 '11 at 22:09

You may consider this question from perspective of energy view: the electric energy is consumed by the resistor and convert this energy to the thermal energy (the source of heat that heat up the water). So from this point of view, if you can assume the 100% electric energy converting to thermal energy, and usually, this assumption is right for resistors since there is no other kind of energy that electric energy can convert to, since this is not a motor or a light bulb. Thus you will have the following equation:

$Heat = I^2Rt = \frac{U^2}{R}t$

So this amount of heat will be absorbed by water and heat the water up, so

$Heat = c_{water}\cdot m\cdot \Delta T$

By solving these two equations, you can get the amount of water ($m_{water}$) you need as

$m_{water} = \frac{U^2 t}{R\cdot c\cdot \Delta T}$

where $R$ can be easily calculated as $R=\frac{\rho L}{\pi r^2}$ for this cylindrical resistor.

So the key point to connect the electricity to thermal dynamics is the conservation of energy so that energy has to convert from one form (electric energy in your case) to another form (thermal energy or heat in your case), and little or no energy is converted into other form such as light.

-
nice 'physics' answer! –  Nic Oct 21 '11 at 21:01
find m from this equation, $$mC_p\Delta T = \frac{V^2}{R}t$$ where,$$R=\frac{\rho l}{\pi r^2}$$