[nlb149]

A system consisting of 73.2g of liquid water at 298K is heated using an immersion heater at a constant pressure of 1.00 bar.  If a current of 2.25 A passes through the 10.0-ohm resistor for 125. s, what is the final temperature of the water?

Any thoughts on how to approach this problem would be appreciated because I'm pretty lost on where to go on this one.  Thanks.

[Yggdrasil]

Do you know how to calculate the amount of work (energy) used by the heater (if not, you should be able to find the equation in your introductory physics textbook)?  If that energy is used to heat the water, what final temp of water will you get?

[nlb149]

I assume you mean that I use IR = V to get the voltage and calculate the electrical work done which i got to come out to 6.33KJ (I don't know if that is right or not).  But even if that is right I still need to calculate the work done by the expanding volume, which should be equal to -Pext (Vf - Vi) since pressure is constant but I don't have the volumes so I'm not sure how to do it.

[Borek]

Try to answer the question as if it was just a simple heat balance.

I doubt they want you to calculate work done by liquid expanding water.

[nlb149]

I guess you were right.  I used q(p) = w done by the heater and ignored the work done by the liquid.  Then, delta H = q(p) = integral(T1,T2) C(p)dT and came up with 310k which, according to the back of the book, is the right answer.

Thanks guys for the help.