Math · Calculus II · Worked example
Solve Newton’s law of cooling
A cup of tea at 90 °C sits in a 20 °C room and cools to 60 °C in 10 minutes. Newton’s law of cooling says dT/dt = −k(T − 20). Find T(t), and find when the tea reaches 30 °C.
Separate and integrate
dT/(T − 20) = −k dt integrates to ln|T − 20| = −kt + C, so T − 20 = Ae^(−kt).
Use the starting temperature
At t = 0 the difference is 70 °C, so A = 70 and T = 20 + 70e^(−kt).
Find k from the second reading
T(10) = 60 gives 70e^(−10k) = 40, so e^(10k) = 7/4.
Find when it reaches 30 °C
70e^(−kt) = 10 gives e^(kt) = 7, so t = (ln 7)/k = 10 ln 7/ln(7/4), about 34.8 minutes.
Check the model
With k ≈ 0.056 per minute, T(10) ≈ 60.0 °C. The model assumes the room stays at 20 °C and the tea loses heat at a rate proportional to the temperature difference.
Result
T(t) = 20 + 70e^(−kt) with k = ln(7/4)/10 ≈ 0.056 per minute; the tea reaches 30 °C after about 34.8 minutes.
Your turn
A metal part at 200 °C cools in 25 °C air and reaches 100 °C after 5 minutes. Write T(t).
Show the answer and explanation
T(t) = 25 + 175e^(−kt), with k = ln(7/3)/5 ≈ 0.169 per minute.
A = 200 − 25 = 175, and 175e^(−5k) = 75 gives e^(5k) = 7/3.
Keep exploring
Derivative & antiderivative checks verifies that T = 20 + 70e^(−0.056t) has derivative −3.92e^(−0.056t), which is −0.056(T − 20).
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