A-Level Differential equations: worked solution

Question

Tea at 85°C is left to cool in a 25°C room; after 20 minutes it is 55°C. Its temperature falls at a rate proportional to the difference between the tea and room temperatures. With θ the tea temperature t minutes after cooling begins, show that θ=25+60e^(−kt) and find the constant k.

Step 1: turn the description into a differential equation. "The temperature falls at a rate proportional to the difference between the tea and the room" means dθ/dt = −k(θ − 25), where k > 0 is the constant of proportionality. The minus sign is there because θ is decreasing while θ − 25 is positive. Step 2: separate the variables and integrate. 1/(θ − 25) dθ = −k dt ∫ 1/(θ − 25) dθ = ∫ −k dt ln|θ − 25| = −kt + c. Step 3: make θ the subject. Taking exponentials of both sides, |θ − 25| = e^(−kt + c) = e^c × e^(−kt). The tea stays hotter than the room, so θ − 25 > 0 and the modulus can be dropped. Writing A = e^c, θ − 25 = A e^(−kt), that is θ = 25 + A e^(−kt). Step 4: use the starting temperature to find A. At t = 0 the tea is 85°C: 85 = 25 + A e⁰ = 25 + A, so A = 60. Hence θ = 25 + 60 e^(−kt), as required. Step 5: use the second reading to find k. At t = 20, θ = 55: 55 = 25 + 60 e^(−20k) 30 = 60 e^(−20k) e^(−20k) = 1/2. Take natural logarithms of both sides: −20k = ln(1/2) = −ln 2 k = (ln 2)/20 = (1/20)ln 2 = 0.034657... k ≈ 0.0347 (3 significant figures).