Question 9.4.2: Temperature Dynamics The temperature of liquid cooling in a ...

We will model the liquid as a single lumped thermal mass with a representative temperature T .
From conservation of heat energy we have
\frac{d  E}{d t} = − \frac{T  −  T_{o}}{R}

where E is the heat energy in the liquid, T_{o} = 68°F is the temperature of the air surrounding the cup, and R is the total thermal resistance of the cup. We have E = m c_{p}(T  −  T_{o}) = C(T  −  T_{o}) , where m is the liquid mass, c_{p} is its specific heat, and C = m c_{p} is the thermal capacitance.
Assuming that m, c_{p}, and T_{o} are constant, we obtain
C \frac{d T}{d t} = − \frac{T  −  T_{o}}{R}
If we let \Delta T = T  −  T_{o} and note that
\frac{d( \Delta T )}{d t} = \frac{d T}{d t}
we obtain
RC \frac{d( \Delta T )}{d t} + \Delta T = 0       (1)
The time constant is \tau = RC, and the solution has the form
\Delta T (t) = \Delta T (0)e^{−t/τ}
Thus,
\ln \Delta T (t) = \ln \Delta T (0)  −  \frac{t}{τ}          (2)
The transformed data \ln \Delta T (t) are plotted in Figure 9.4.2a. Because they fall near a straight line, we can use equation (2) to fit the data. The values obtained are \ln T (0) = 4.35 and \tau = 2792  sec. This gives T (0) = 77°F. Thus the model is
T (t) = 68 + 77e^{−t/2792}              (3)
The computed time to reach 120°F is
t = −2792 \ln \frac{120  −  68}{77} = 1112  sec
The plot of equation (3), along with the data and the estimated point (1112, 120) marked with a “+” sign, is shown in part (b) of Figure 9.4.2. Because the graph of our model lies near the data points, we can treat its prediction of 1112 sec with some confidence.