THERMAL BYU 14

Concept: The time taken for cooling depends on the ratio of heat capacity (Heat Content) to the rate of heat loss.

Let the characteristic linear dimension be $L$.

• Heat Content ($Q$): $Q = ms\Delta T$. Since mass $m = \rho V$ and $V \propto L^3$: $$Q \propto L^3$$
• Rate of Heat Loss ($R$): According to Newton’s law of cooling, $R \propto \text{Area}$. Since Area $A \propto L^2$: $$R \propto L^2$$

The cooling time interval $\Delta t$ is given by: $$\Delta t \approx \frac{Q}{R} \propto \frac{L^3}{L^2} \propto L$$

Calculation:
We are given that the second tub has linear dimensions $\eta = 8$ times the first one ($L_2 = 8 L_1$). $$\frac{t_2}{t_1} = \frac{L_2}{L_1} = \eta$$ $$t_2 = \eta t_1$$ Given $t_1 = 20$ min and $\eta = 8$: $$t_2 = 8 \times 20 = 160 \text{ min}$$