A body at 80°C is placed in a room kept at 20°C. Its temperature T (in °C) after t minutes satisfies \(\displaystyle\frac{dT}{dt} = -k(T - 20)\), where k is a positive constant. If the temperature of the body is 50°C after 10 minutes, then its temperature after 20 minutes is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 35°C
The variables separate:
\( \displaystyle \frac{dT}{T - 20} = -k\,dt \)
Integrating both sides (T stays above 20):
\(\log(T - 20) = -kt + c\)
\(T - 20 = Ae^{-kt}\)
At t = 0, T = 80:
\(A = 60\)
\(T - 20 = 60e^{-kt}\)
At t = 10, T = 50:
\(30 = 60e^{-10k}\)
\( \displaystyle e^{-10k} = \frac{1}{2} \)
At t = 20, \(e^{-20k} = \left(e^{-10k}\right)^2 = \frac{1}{4}\):
\( \displaystyle T - 20 = 60 \times \frac{1}{4} = 15 \)
\(T = 35\)
The excess over room temperature halves every 10 minutes (60, 30, 15); the temperature itself does not fall by equal steps, which is why 20°C is wrong.