QuestionDecember 16, 2025

100. A cup of hot chocolate has temperature 80^circ C in a room kept at 20^circ C After half an hour the hot chocolate cools to 60^circ C a. What is the temperature of the chocolate after another half hour? b. When will the chocolate have cooled to 40^circ C

100. A cup of hot chocolate has temperature 80^circ C in a room kept at 20^circ C After half an hour the hot chocolate cools to 60^circ C a. What is the temperature of the chocolate after another half hour? b. When will the chocolate have cooled to 40^circ C
100. A cup of hot chocolate has temperature 80^circ C in a room kept at 20^circ C After half an hour the hot chocolate cools to 60^circ C a. What is the temperature of the chocolate after another half hour? b. When will the chocolate have cooled to 40^circ C

Solution
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Answer

a. \approx 26.67^\circ C after another half hour. ### b. 0.5 hours (30 minutes) to cool to 40^\circ C. Explanation 1. Write Newton’s Law of Cooling T(t) = T_a + (T_0 - T_a)e^{-kt}, where T_a=20^\circ C, T_0=80^\circ C. 2. Find cooling constant k At t=0.5 hr, T(0.5)=60: 60 = 20 + 60e^{-0.5k} \implies e^{-0.5k} = \frac{2}{6} = \frac{1}{3} \implies -0.5k = \ln\left(\frac{1}{3}\right) \implies k = -2\ln\left(\frac{1}{3}\right) = 2\ln 3. 3. Find temperature at t=1 hr T(1) = 20 + 60e^{-k}; e^{-k} = (e^{-0.5k})^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}; so T(1) = 20 + 60 \times \frac{1}{9} = 20 + \frac{60}{9} = 20 + \frac{20}{3} = \frac{80}{3} \approx 26.67^\circ C. 4. Solve for time when T(t) = 40^\circ C 40 = 20 + 60e^{-kt} \implies 20 = 60e^{-kt} \implies e^{-kt} = \frac{1}{3}. Since e^{-0.5k} = \frac{1}{3}, -kt = -0.5k \implies t = 0.5 hr.

Explanation

1. Write Newton’s Law of Cooling<br /> $T(t) = T_a + (T_0 - T_a)e^{-kt}$, where $T_a=20^\circ C$, $T_0=80^\circ C$.<br />2. Find cooling constant $k$<br /> At $t=0.5$ hr, $T(0.5)=60$: $60 = 20 + 60e^{-0.5k} \implies e^{-0.5k} = \frac{2}{6} = \frac{1}{3} \implies -0.5k = \ln\left(\frac{1}{3}\right) \implies k = -2\ln\left(\frac{1}{3}\right) = 2\ln 3$.<br />3. Find temperature at $t=1$ hr<br /> $T(1) = 20 + 60e^{-k}$; $e^{-k} = (e^{-0.5k})^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$; so $T(1) = 20 + 60 \times \frac{1}{9} = 20 + \frac{60}{9} = 20 + \frac{20}{3} = \frac{80}{3} \approx 26.67^\circ C$.<br />4. Solve for time when $T(t) = 40^\circ C$<br /> $40 = 20 + 60e^{-kt} \implies 20 = 60e^{-kt} \implies e^{-kt} = \frac{1}{3}$. Since $e^{-0.5k} = \frac{1}{3}$, $-kt = -0.5k \implies t = 0.5$ hr.
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