10.1 The Cooling Coffee Problem Revisited

• In this section we revisit the cooling coffee cup problem from Chapters 9.1 and 9.2, where an ODE was derived using Newton's law of cooling.
• We will find the analytic solution using separation of variables, and numerically using RK4 in R.
  
• We will also solve an ODE that uses the natural law of cooling, and compare the two solutions.
  

• Assume air is moving over and around cup.

From Ch9.2:

\[ \frac{dU}{dt} = -\lambda (U-u_s), \,\, \lambda= \frac{hS}{cm} \]

Parameters:

• \( c \) = specific heat of coffee
• \( m \) = mass of coffee
• \( h \) = Newton cooling coefficient
• \( S \) = surface area through which heat is lost
• \( u_s \) = constant temperature of surroundings

Use separation of variables with \( U(0)=u_{0} \) and \( U > u_{s} \):

\[ \small{ \begin{aligned} \frac{dU}{dt} &=-\lambda(U-u_{s}) \\ \int \frac{dU}{U-u_{s}} & =-\lambda \int dt \\ \ln(U-u_{s}) &=-\lambda t +K \\ U-u_{s} &= e^{-\lambda t + K} \\ U &= e^{-\lambda t}e^{K} + u_{s} \end{aligned}} \]

Applying the initial condition \( U(0)=u_{0} \) we have

\[ U(t) = (u_{0}-u_{s})e^{-\lambda t}+u_{s} \]

If it takes 10 minutes for a cup of coffee to cool from \( 60^\circ \) C to \( 50 ^\circ \) C in a room at temperature \( 20^\circ \) C, how long does it take to cool from \( 60^\circ \) C to \( 40 ^\circ \) C?

First, find \( \lambda \): \[ \small{ \begin{aligned} U(t) & = (u_{0}-u_{s})e^{-\lambda t}+u_{s} \\ U - u_s & = (u_{0}-u_{s})e^{-\lambda t} \\ e^{-\lambda t} & = \frac{U - u_s}{u_{0}-u_{s}} \\ -\lambda t & = \ln \left(\frac{U - u_s}{u_{0}-u_{s}} \right) \\ \lambda &= -\frac{1}{t} \ln \left(\frac{U - u_s}{u_{0}-u_{s}} \right) \\ \end{aligned}} \]

Recall that it takes 10 minutes for a cup of coffee to cool from \( 60^\circ \) C to \( 50 ^\circ \) C in a room at temperature \( 20^\circ \) C.

Thus we substitute \( U=50, u_0 = 60, u_s = 20, t = 10 \) into our equation and to find \( \lambda \):

\[ \small{ \lambda = \frac{1}{t} \ln \left(\frac{U - u_s}{u_0 - u_s} \right) = -\frac{1}{10} \ln \left(\frac{50 - 20}{60-20} \right) \cong 0.0288 } \]

-1/10*log(30/40)

[1] 0.02876821

Now that we know what \( \lambda \) is, we adapt above formula to solve for time \( T \) required for coffee to cool from \( 60^\circ \) C to \( 40 ^\circ \) C:

\[ \small{ \begin{aligned} T &= \frac{1}{\lambda} \ln \left(\frac{U - u_s}{u_0 - u_s} \right) \\ &= -\frac{1}{0.0288} \ln \left(\frac{40 - 20}{60-20} \right) \\ & \cong 24.1 \end{aligned}} \]

-1/0.0288*log(20/40)

[1] 24.06761

• Assume air is motionless over and around cup.

\[ \frac{dU}{dt}=-\lambda(U-u_s)^{5/4} \]

Apply the law of natural cooling to obtain

\[ \frac{dU}{dt} = -\lambda (U-u_s)^{\frac{5}{4}}, \,\, \lambda =\frac{h_1S}{cm} \]

This differential equation is separable and can be solved using the same technique as before (Exercise 10.1).

For specifications as in Example 2, it takes 25.3 minutes to reach \( 40° \) C, which is slower than the 24 minutes in Example 2.

• The estimation time using the natural cooling law is slightly longer than that obtained using Newton's law of cooling.
• This makes sense because we expect cooling with a slight breeze (Newton's law of cooling) to be more effective than cooling in perfectly still air (natural law of cooling).

• Newton's law of cooling

\[ \frac{dU}{dt} = -\lambda (U-u_s), \,\, \lambda= \frac{hS}{cm} \]

• Natural law of cooling

\[ \frac{dU}{dt} = -\lambda (U-u_s)^{\frac{5}{4}}, \,\, \lambda =\frac{h_1S}{cm} \]