Heat engine

Here's a code-based solution:

import math

C = 4184.0 # J/(kg * K)

T1 = 100.0 + 273.15
T2 = 20.0 + 273.15

m1 = 0.4
m2 = 2.0

Qheat = m1*C*(T1 - T2)

dQ = 10.0**(-3.0)

W = 0.0

while T1 - T2 > 0.001:

    dT1 = dQ / (m1*C)

    T1 = T1 - dT1

    e = 1.0 - T2/T1

    dW = dQ * e

    W = W + dW

    dQ2 = dQ * (1.0 - e)

    dT2 = dQ2 / (m2*C)

    T2 = T2 + dT2

print T1
print T2
print ""
print W/Qheat

The optimal thermodynamic efficiency (Carnot efficency) reads $\frac{dW}{|dQ_1|} = \eta = 1 - \frac{T_2}{T_1}$ With initial temperatures $T_1 = 373 \,\text{K}$ and $T_2 = 293 \,\text{K}$ , the heat engine start with an efficiency of $\eta \approx 21 \,\text{\%}$ . However, the efficiency decreases in the course of the heat transfer, which adjusts the temperature of the coffee and the water. With the help of energy conservation $-dQ_1 = dW + dQ_2$ , we get $\begin{aligned} & & - \frac{C_2 dT_2}{C_1 d T_1} &= -\frac{dQ_2}{dQ_1} = 1 - \eta = \frac{T_2}{T_1} \\ \Rightarrow & & C_2 \frac{dT_2}{T_2} &= - C_1 \frac{dT_1}{T_1} \\ \Rightarrow & & C_2 \int_{T_2}^{T_f} \frac{dT_2}{T_2} &= - C_1 \int_{T_2}^{T_f} \frac{dT_1}{T_1} \\ \Rightarrow & & C_2 \ln \frac{T_f}{T_2} &= - C_1 \ln \frac{T_f}{T_1} \\ \Rightarrow & & \left( \frac{T_f}{T_2} \right)^{C_2} &= \left( \frac{T_f}{T_1} \right)^{-C_1} \\ \Rightarrow & & T_f &= T_1^x T_2^{1-x} \approx 305.18 \,\text{K}, \quad x = \frac{C_1}{C_1 + C_2} = \frac{1}{6} \end{aligned}$ This final temperature $T_f$ is slightly lower than the weighted average $T_m = \frac{C_1 T_1 + C_2 T_2}{C_1 + C_2} \approx 306.48 \,\text{K}$ that would result as mixing temperature from coffee and water. The energy difference $W = (C_1 + C_2) T_m - (C_1 + C_2) T_f$ corresponds to the work $W$ of the heat engine. To heat the coffee water was an energy of $Q_\text{heat} = C_1 (T_1 - T_2)$ necessary, so that the recovery efficiency results to $\eta_\text{rec} = \frac{W}{Q_\text{heat}} = 6 \frac{T_m - T_f}{T_1 - T_2} \approx 9.78 \,\text{\%}$