A body cools from \( 80^{\circ}C \) to \( 50^{\circ}C \) in 5 min. Calculate the time it takes to cool from \( 60^{\circ}C \) to \( 30^{\circ}C \) if the surrounding temperature is \( 20^{\circ}C \)
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Using the average temperature \( \frac{\theta_1 + \theta_2}{2} \) as an approximation for the body's temperature during the interval is a highly efficient way to solve cooling problems without using calculus integration.
Step 1: State Newton's law of cooling. A hot body cools faster when it is much hotter than its surroundings. The simple working form uses the average temperature of the body during the interval: \[ \frac{\theta_1 - \theta_2}{t} = K\left[\frac{\theta_1 + \theta_2}{2} - \theta_s\right] \] Step 2: Apply it to the first cooling. The body goes from $80^{\circ}$C to $50^{\circ}$C in $5$ minutes with surroundings at $20^{\circ}$C: \[ \frac{80 - 50}{5} = K\left[\frac{80 + 50}{2} - 20\right] \] Step 3: Find the cooling constant K. \[ 6 = K[65 - 20] = 45K \;\Rightarrow\; K = \frac{6}{45} = \frac{2}{15} \] Step 4: Apply it to the second cooling. Now the body goes from $60^{\circ}$C to $30^{\circ}$C in unknown time $t$: \[ \frac{60 - 30}{t} = K\left[\frac{60 + 30}{2} - 20\right] = K[45 - 20] \] So $\dfrac{30}{t} = 25K$. Step 5: Put the value of K in. \[ \frac{30}{t} = 25\left(\frac{2}{15}\right) = \frac{10}{3} \] Step 6: Solve for the time. \[ 10t = 90 \;\Rightarrow\; \boxed{t = 9\ \text{min}} \]