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I'm a first time user and I hope I won't be too enigmatic asking the following question:

I have a cylinder (radius= $6$ $\text{cm}$) with a frictionless piston on top of it and inside $30$ $g$ of $H_2 O$ (molar weight is $M=30$). External pressure is $1 \text{ atm}$ and initial temperature is $T_1=273$ $K$. $Q=24$ $KJ$ of heat are put into the system. What height does the piston reach?

I'm pretty sure I didn't get this right, but here is what I did:

$$Q = Q_\text{water} - Q_\text{Latent evaporation heat} + Q_\text{steam}= mc \times (373-273) - m L + (m/M)Cv \times (T_2-373)$$

from this equation I found $T_2$ and knowing that pressure is constant, I found the volume of the steam and thus the height of the piston which resulted something crazy like $27 \text{m}$. What have I done wrong?

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I would work in kg rather than moles since all the thermodynamic data is given as J/kg. I had a quick stab at the calculation and got a number between 0.5 and 1m (I won't spoil your fun by giving the exact number :-). If you give detailed working I will try and see what point you go wrong. I would start by working out how much of the 24kJ is used to heat the water to 373K, then work out how much of the water can be vaporised with the remaining energy. –  John Rennie Feb 21 '13 at 12:47
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