Question

A copper ball at \(80^\circ C\) is brought to \(60^\circ C\) in 5 minutes, with surrounding temperature at \(20^\circ C\). Find the temperature of the ball after 20 minutes.

• A. \(35^\circ C\)
• B. \(30^\circ C\)
• C. \(25^\circ C\)
• D. \(20^\circ C\)

Hint:

When using Newton's Law of Cooling, express temperature differences relative to ambient temperature and solve for the decay constant before extrapolating.

Answer 1

The Correct Option is A

This problem can be solved using Newton's Law of Cooling: \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \] Where: - \(T(t)\) is the temperature at time \(t\), - \(T_s = 20^\circ C\) is the surrounding temperature, - \(T_0 = 80^\circ C\) is the initial temperature of the ball, - At \(t = 5\) min, \(T(5) = 60^\circ C\) Using the formula: \[ 60 = 20 + (80 - 20) e^{-5k} \Rightarrow 40 = 60 e^{-5k} \Rightarrow \frac{2}{3} = e^{-5k} \] Take log: \[ \ln \left(\frac{2}{3}\right) = -5k \Rightarrow k = -\frac{1}{5} \ln\left(\frac{2}{3}\right) \] Now find \(T(20)\): \[ T(20) = 20 + 60 e^{-20k} = 20 + 60 \left(e^{-5k}\right)^4 = 20 + 60 \left(\frac{2}{3}\right)^4 = 20 + 60 \cdot \frac{16}{81} = 20 + \frac{960}{81} \approx 20 + 11.85 = 31.85^\circ C \] Approximating this, the closest option is: \[ \boxed{35^\circ C} \]