Question

一道积分题

172°C的咖啡放进一个杯子，留在77°c的屋子里，5分钟后，咖啡变成138°C。假设dT/dt=k(T-77)。1.求15分钟后咖啡的温度。
2.多少分钟后100°C
列出式子即可~

Answer 1

dT/dt = k(T-77), dT/(T-77) = kdt， 两边积分得
ln(T-77) = kt + lnC, T-77 = Ce^(kt)
t = 0 , T = 172, 代入T-77 = Ce^(kt) 得 95 = C，
t = 5 , T = 138, 代入T-77 = Ce^(kt) 得 61 = Ce^(5k) = 95e^(5k)，
则 e^(5k) = 61/95, k = (1/5)ln(61/95)，
T = 77+ 95e^[(1/5)ln(61/95)t]
(1) t = 15时, T = 77+ 95e^[3ln(61/95)] = 77+ 95(61/95)^3 ≈ 102.15°C
(2) 100 = 77+ 95e^[(1/5)ln(61/95)t], e^[(1/5)ln(61/95)t] = 23/95
[(1/5)ln(61/95)t] = ln(23/95), t = 5ln(23/95)/ln(61/95) = ...

Answer 2

(1)
dT/dt=k(T-77)
∫dT/(T-77) = ∫kdt
ln|T-77| = kt + C
T(0)=172
ln|T-77| = kt + C
ln|172-77| = 0+C
C=ln95
ln|T-77| = kt + ln95
T-77 = 95e^(kt)
T = 77 +95e^(kt)
T(5) = 138°C
138=77 +95e^(5k)
95e^(5k) =61
k =-(1/5)ln(95/61)
T = 77 +95e^[-(1/5)t. ln(95/61) ]
15分钟后咖啡的温度
=T(15)
= 77 +95e^[-(1/5)(15)ln(95/61) ]
= 77 +95e^[-3ln(95/61) ]
=77 + 95.( 61/95)^3
=77.0028°C
(2)
多少分钟后100°C
T = 77 +95e^[-(1/5)t. ln(95/61) ]
100=77 +95e^[-(1/5)t. ln(95/61) ]
95e^[-(1/5)t. ln(95/61) ] =23
-(1/5)t. ln(95/61) = ln(23/95)
t = -5ln(23/95)/ln(95/61)
=3.14 min