What is the obtainable work from cooling water? In Callen's thermostatistics textbook, problem 4.5-14 says 

If the temperature of the atmosphere is 5°C on a winter day and if 1 kg of water at 90°C is available, how much work can be obtained as the water is cooled to the ambient temperature? Assume that the volume of the water is constant, and assume that the molar heat capacity at constant volume is 75 J/mole K and is independent of temperature.

This seems straightforward enough, but I am not getting the right answer, and I am not sure why. 
I first converted 1 kg of water to moles: 1 kg = 1000 g, water is 18g/mol so 1000/18=number of moles of water=55.55 moles. Then the temperature difference is 90-5=85. So then the heat is 75 * 55.55 * 85 = 354131.25 J but the textbook says the answer is 45 * 10^3 J? 
 A: So I figured it out myself: 
For a constant volume process, $\Delta S$ is given by $\Delta S=\int \frac{N C_{v}}{T}dT=N C_{v} \ln \left(\frac{T_{2}}{T_{1}}\right)$. 
1 kg of water has 55.55 moles of water. Water is 18g/mol and 1 kg is 1000 g. So 1000 / 18 = number of moles = 55.55 moles. $T_i=363.15\;\mathrm{K}$ (90 deg C) and $T_f=278.15\;\mathrm{K}$ (5 deg C). Volume is kept constant. 
$d S_{\text {total }}=d S_{\text {primary }}+d Q / T_{r}$ where $d S_{\text {primary }}=N C_{v} \ln \left(\frac{T_{f}}{T_{i}}\right)$, $T_r=278.15\;\mathrm{K}$, $dQ$ is found given $d S_{\text {total }}=0$. We have $d Q=-d S_{\text {primary }} * T_{r}=-278.15 * 55.55 * 75 * \ln \left(\frac{278.15}{363.15}\right)=309011.6 \;\mathrm{J}$. 
Also $d U=d Q+d W$ and $d U=\Delta U= \int N C_{v} d T=N C_{v} \Delta T=55.55 * 75 * 85=354131.3\; \mathrm{J}$. 
So $d W=d U-d Q=354131.3-309011.6=\boxed{45119.6 \;\mathrm{J}}$ as desired. 
The shortcut formula $d W=d U+T_{r} d S_{p r i m a r y}-T_{r} d S_{t o t a l}$ after substituting $dQ$ and $d S_{\text {total }}=0$ can also be used. 
