Curd is at 80° F, five minutes later it came down at 60°F. After another 5 minutes, its temperature became 50° F. Given that the rate of change of temperature is proportional to (T - S), where S is temperature of surroundings and T is temperature of the curd at any time t. Then the temperature of the surroundings is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 40°F
This is Newton’s law of cooling. With k a positive constant:
\( \displaystyle \frac{dT}{dt} = -k(T - S) \)
Separating the variables and integrating:
\( \displaystyle \int \frac{dT}{T - S} = -\int k\,dt \)
\(\log|T - S| = -kt + c\)
\(T - S = (80 - S)e^{-kt}\)
(using T = 80 at t = 0). At t = 5 and t = 10:
\(60 - S = (80 - S)e^{-5k}\)
\(50 - S = (80 - S)e^{-10k}\)
Squaring the first equation and using the second, since \(e^{-10k} = (e^{-5k})^2\):
\( \displaystyle (60 - S)^2 = (80 - S)^2 e^{-10k} \)
\( \displaystyle (60 - S)^2 = (80 - S)(50 - S) \)
\(3600 - 120S + S^2 = 4000 - 130S + S^2\)
\(10S = 400\)
\(S = 40\)
So the surroundings are at 40°F.
Tip: For readings taken at equal time intervals, the excesses over the surroundings, T − S, are in geometric progression: here 40, 20, 10 for S = 40.