# Entropy and heat

by Benzoate
Tags: entropy, heat
 Sci Advisor HW Helper PF Gold P: 2,532 $\Delta S = \frac{Q}{T}$ isn't going to give the right answer, because T isn't constant. Try integrating $$dS=c\,m\,\frac{dT}{T}[/itex] and plugging the initial and final temperatures into the resulting equation.  P: 15 You simply have to integrate (sum over the transformation path) [tex]dS=\frac{dU}{T} + \frac{PdV}{T} - \sum_i \frac{\mu_i dN_i}{T}$$ Fourier's law tells you that $$dU=\rho c dT$$ hence what is written above. You could also ask yourself what has happened with the second term (the last one is irrelevant here) $$\frac{PdV}{T}$$ Hint: linked to the ice coefficient of thermal expansion $$\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$$