The temperature of a chemical reaction mixture, H ∘CH\, ^\circ\text{C}H∘C, is modelled by the equation
H=15−e0.5t−3−12e−0.4t,t≥0 H = 15 - \text{e}^{0.5t-3} - 12\text{e}^{-0.4t}, \quad t \ge 0 H=15−e0.5t−3−12e−0.4t,t≥0where ttt is the time in minutes after the reaction is initiated. A graph of the temperature against time shows that it starts at a positive value, rises to a peak, and then gradually falls.
Using calculus, determine the maximum temperature of the mixture during this reaction phase.
The accumulated thermal index, SSS, is defined by the integral of the temperature over a time interval. Given that the index reaches a value of 65 at time TTT, such that ∫0TH dt=65\int_{0}^{T} H \, \text{d}t = 65∫0THdt=65,
show that TTT is a solution of the equation
T=115(95−30e−0.4T+2e0.5T−3−2e−3) T = \frac{1}{15} (95 - 30\text{e}^{-0.4T} + 2\text{e}^{0.5T-3} - 2\text{e}^{-3}) T=151(95−30e−0.4T+2e0.5T−3−2e−3)Using the iteration formula
Tn+1=115(95−30e−0.4Tn+2e0.5Tn−3−2e−3) T_{n+1} = \frac{1}{15} (95 - 30\text{e}^{-0.4T_n} + 2\text{e}^{0.5T_n-3} - 2\text{e}^{-3}) Tn+1=151(95−30e−0.4Tn+2e0.5Tn−3−2e−3)with T1=6T_1 = 6T1=6, find to 4 decimal places:
(i) the value of T2T_2T2,
(ii) the time taken for the accumulated thermal index to reach 65.
