Problem — Newton's cooling

Problem

A hot body at $80 \ ^{\circ}\text{C}$ is placed in a room at $20 \ ^{\circ}\text{C}$. The temperature is measured over time, giving the following table:

| t (min) | 0 | 5 | 10 | 20 | 40 | 80 | | T (°C) | 80 | 58 | 44 | 30 | 22,5 | 20,5 |

(a) Verify qualitatively that the cooling follows Newton’s law, $T(t) - T_{amb} = (T_0 - T_{amb})\,e^{-kt}$, with $T_0 = 80 \ ^{\circ}\text{C}$ and $T_{amb} = 20 \ ^{\circ}\text{C}$. (b) Estimate the constant $k$.

Measured temperature (blue points) and interpolating exponential curve (red) with the room temperature dashed.

Solution

It is convenient to reason in terms of the temperature excess above ambient, $\theta = T - 20$: this gives the values $\theta(0) = 60$, $\theta(5) = 38$, $\theta(10) = 24$, $\theta(20) = 10$, $\theta(40) = 2{,}5$, $\theta(80) = 0{,}5$ (in degrees). (a) Over equal time intervals, the ratios between successive values of the excess are roughly constant (for example $38/60 \approx 0{,}63$ and $24/38 \approx 0{,}63$): this is the typical sign of an exponential decay, so Newton’s law is verified. (b) From the relation $\theta = 60 \, e^{-kt}$ we can obtain $k$ by inverting the exponential at various instants: $\begin{aligned} t = 5: \ & k = -\frac{\ln(38/60)}{5} \approx 0{,}091 \\ t = 10: \ & k = -\frac{\ln(24/60)}{10} \approx 0{,}092 \\ t = 20: \ & k = -\frac{\ln(10/60)}{20} \approx 0{,}090 \end{aligned}$ The three values are consistent; the average is $k \approx 0{,}092 \ \text{min}^{-1}$. The half-life of the temperature excess is $t_{1/2} = \dfrac{\ln 2}{k} \approx \dfrac{0{,}693}{0{,}092} \approx 7{,}6 \ \text{min}$. $\ev{k \approx 0{,}092 \ \text{min}^{-1} \qquad t_{1/2} \approx 7{,}6 \ \text{min}}$

Collegamenti

Argomenti: Termologia Concetti: Trasmissione del calore