7FNewton


 * Newton's Law of Cooling**

The rate at which a body cools (temperature decreases) is proportional to the difference between the temperature of the body and the surroundings. ... ... {Body is used to describe any object that is cooling}


 * Let T(t) = the temperature of the body at time, t.
 * Let Ts = the temperature of the surroundings (constant)
 * The difference between them is T(t) – Ts
 * (Temperature decreasing so derivative is negative)

math . \qquad \dfrac{dT}{dt} \propto -\left( T - T_S \right) math so math . \qquad \dfrac{dT}{dt} = -k \left( T - T_S \right) math

Solution is an exponential graph with a horizontal asymptote at Ts.

Example

A pie with an initial temperature of 80C is taken from the oven and left to cool. The room temperature is constant at 25C. After 15 minutes, the temperature of the pie is 60C. Assume Newton's Law of Cooling. Find the temperature of the pie at time, t minutes after being taken from the oven. Hence find the temperature of the pie after 30 minutes.

From the question ... ... T 0 = 80C ... ... Ts = 25C ... ... T(15) = 60C ... ... T(30) = ?

Use Newton's Law of Cooling math , \qquad \dfrac{dT}{dt} = -k \left( T - 25 \right) math

Take the reciprocal math . \qquad \dfrac{dt}{dT} = \dfrac{-1}{ k \left( T - 25 \right)}, \; T \neq 25 math

Integrate math . \qquad t=\displaystyle{\int} \dfrac{-1}{ k \left( T - 25 \right)} \, dT math . math . \qquad t=-\dfrac{1}{k} \displaystyle{\int} \dfrac{1}{\left( T - 25 \right)} \, dT math . math \\ . \qquad t=-\dfrac{1}{k} \log_e \left( T - 25 \right)+c \qquad \lbrace \times (-k) \rbrace \\. \\ . \qquad -kt = \log_e \left( T - 25 \right) - kc \\. \\ . \qquad \log_e \left( T - 25 \right) = kc - kt math

Use log laws math \\ . \qquad T - 25 = e^{kc-kt} \\. \\ . \qquad T - 25 = e^{kc} \times e^{-kt} math . math . \qquad \text{Let } A=e^{kc} \; \; \textit{constant} math . math . \qquad T(t) = Ae^{-kt}+25 math

Now find A, using T(0)=80 math \\ . \qquad 80 = Ae^0+25 \\. \\ . \qquad A = 55 math

Now find k, using T(15) = 60 math \\ . \qquad 60 = 55e^{-15k} + 25 \\. \\ . \qquad 35 = 55e^{-15k} math . math . \qquad e^{-15k} = \dfrac{35}{55} = \dfrac{7}{11} math

Use log laws math \\ . \qquad -15k = log_e \left( \dfrac{7}{11} \right) \\. \\ . \qquad k = \dfrac{-1}{15}log_e \left( \dfrac{7}{11} \right) \\. \\ . \qquad k = 0.030 math ​ **Thus equation is:** math . \qquad T(t) = 55e^{-0.03t} + 25, \quad t \geqslant 0 math

Now find T(30) math \\ . \qquad T(30) = 55e^{-0.03 \times 30} + 25 \\. \\ . \qquad T(30) \approx 47.4^{\circ}C math

{Notice that because t > 0, the graph is restricted to the 1st Quadrant}

General Solution to Newton's Law of Cooling

For Newton's Law of Cooling: math . \qquad \dfrac{dT}{dt} = -k \left( T - T_S \right) math

the general solution will be math . \qquad T(t) = \left( T_0 - T_S \right) e^{-kt} + T_S math

**See also:** www.mathsonline.com.au Y12Extension --> Calculus --> Physical Applications --> Lesson 1 .