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IAL 2025 May Q6

A Level / Edexcel / P3

IAL 2025 May Paper · Question 6

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The temperature, θ∘C\theta^\circ\text{C}, of a computer processor, tt minutes after the computer is switched off, is modelled by the equation

θ=21+Ae−kt\theta=21+Ae^{-kt}

where AA and kk are positive constants.

Given that the temperature of the processor was 75∘C75^\circ\text{C} when the computer was switched off,

(a) find the value of AA.

(2)

Given also that it takes 5 minutes for the temperature of the processor to decrease from 75∘C75^\circ\text{C} to 25∘C25^\circ\text{C},

(b) find the value of kk, giving your answer to 3 significant figures.

(3)

At time TT minutes, the temperature of the processor is decreasing at a rate of 9∘C9^\circ\text{C} per minute.

(c) Find the value of TT according to the model, giving your answer to 2 decimal places.

(3)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

计算机关机后 tt 分钟时，处理器温度 θ∘C\theta^\circ\text{C} 满足模型

θ=21+Ae−kt\theta=21+Ae^{-kt}

其中 A,kA,k 为正常数。

已知关机时处理器温度为 75∘C75^\circ\text{C}，

(a) 求 AA 的值；

又已知处理器温度从 75∘C75^\circ\text{C} 下降到 25∘C25^\circ\text{C} 需要 5 分钟，

(b) 求 kk 的值，答案保留 3 位有效数字；

在 TT 分钟时，处理器温度正以每分钟 9∘C9^\circ\text{C} 的速率下降。

(c) 根据模型求 TT 的值，答案保留到小数点后 2 位。

解答

(a)

When the computer was switched off,

t=0,θ=75t=0,\qquad \theta=75

Substitute into

θ=21+Ae−kt\theta=21+Ae^{-kt}

to get

75=21+Ae075=21+Ae^0

Since e0=1e^0=1,

75=21+A75=21+A

Therefore

A=54\boxed{A=54}

(b)

Using A=54A=54,

θ=21+54e−kt\theta=21+54e^{-kt}

After 5 minutes, θ=25\theta=25, so

25=21+54e−5k25=21+54e^{-5k}

Then

54e−5k=454e^{-5k}=4

and

e−5k=454=227e^{-5k}=\frac{4}{54}=\frac{2}{27}

Take natural logs:

−5k=ln⁡(227)-5k=\ln\left(\frac{2}{27}\right)

Thus

k=−15ln⁡(227)=0.520537…k=-\frac15\ln\left(\frac{2}{27}\right)=0.520537\ldots

So, to 3 significant figures,

k=0.521\boxed{k=0.521}

(c)

Differentiate

θ=21+54e−kt\theta=21+54e^{-kt}

with respect to tt:

dθdt=−54ke−kt\frac{\mathrm{d}\theta}{\mathrm{d}t} =-54ke^{-kt}

Using

k=0.520537…k=0.520537\ldots

we get

dθdt=−28.109…e−0.520537…t\frac{\mathrm{d}\theta}{\mathrm{d}t} =-28.109\ldots e^{-0.520537\ldots t}

At time TT, the temperature is decreasing at 9∘C9^\circ\text{C} per minute, so

dθdt=−9\frac{\mathrm{d}\theta}{\mathrm{d}t}=-9

Therefore

−28.109…e−0.520537…T=−9-28.109\ldots e^{-0.520537\ldots T}=-9

So

e−0.520537…T=928.109…e^{-0.520537\ldots T}=\frac{9}{28.109\ldots}

Taking natural logs,

−0.520537…T=ln⁡(928.109…)-0.520537\ldots T =\ln\left(\frac{9}{28.109\ldots}\right)

Hence

T=2.187…T=2.187\ldots

So, to 2 decimal places,

T=2.19\boxed{T=2.19}