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IAL 2021 Jan Q5

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 5

题目

Problem

The temperature, θ ∘C\theta\ ^\circ\text{C}, inside an oven, tt minutes after the oven is switched on, is given by

θ=A−180e−kt\theta=A-180e^{-kt}

where AA and kk are positive constants.

Given that the temperature inside the oven is initially 18∘C18^\circ\text{C},

(a) find the value of AA.

(2)

The temperature inside the oven, 5 minutes after the oven is switched on, is 90∘C90^\circ\text{C}.

(b) Show that k=pln⁡qk=p\ln q where pp and qq are rational numbers to be found.

(4)

Hence find

(c) the temperature inside the oven 9 minutes after the oven is switched on, giving your answer to 3 significant figures,

(2)

(d) the rate of increase of the temperature inside the oven 9 minutes after the oven is switched on. Give your answer in  ∘C min−1\ ^\circ\text{C min}^{-1} to 3 significant figures.

(3)
题目中文翻译

烤箱开启后 tt 分钟时，烤箱内部温度 θ ∘C\theta\ ^\circ\text{C} 满足

θ=A−180e−kt\theta=A-180e^{-kt}

其中 AA 和 kk 为正常数。

已知烤箱初始内部温度为 18∘C18^\circ\text{C}，

(a) 求 AA 的值。

已知烤箱开启 5 分钟后内部温度为 90∘C90^\circ\text{C}。

(b) 证明 k=pln⁡qk=p\ln q，其中 pp 和 qq 为待求有理数。

由此求

(c) 烤箱开启 9 分钟后内部温度，答案保留 3 位有效数字；

(d) 烤箱开启 9 分钟后内部温度的升高速率，答案用  ∘C min−1\ ^\circ\text{C min}^{-1} 表示，并保留 3 位有效数字。

解答

(a)

The temperature model is

θ=A−180e−kt\theta=A-180e^{-kt}

Initially,

t=0,θ=18t=0,\qquad \theta=18

Substitute these values:

18=A−180e018=A-180e^0

Since

e0=1e^0=1

we get

18=A−18018=A-180

Therefore

A=198\boxed{A=198}

(b)

When

t=5t=5

the temperature is

θ=90\theta=90

Using

A=198A=198

we get

90=198−180e−5k90=198-180e^{-5k}

Rearrange:

180e−5k=108180e^{-5k}=108

So

e−5k=108180=35e^{-5k}=\frac{108}{180}=\frac35

Take natural logarithms:

−5k=ln⁡(35)-5k=\ln\left(\frac35\right)

Therefore

k=−15ln⁡(35)k=-\frac15\ln\left(\frac35\right)

Using log laws,

−ln⁡(35)=ln⁡(53)-\ln\left(\frac35\right)=\ln\left(\frac53\right)

so

k=15ln⁡(53)\boxed{k=\frac15\ln\left(\frac53\right)}

Hence

p=15,q=53\boxed{p=\frac15,\qquad q=\frac53}

(c)

At

t=9t=9

we have

θ=198−180e−9k\theta=198-180e^{-9k}

Using

k=15ln⁡(53)k=\frac15\ln\left(\frac53\right)

gives

θ=198−180e−95ln⁡(53)\theta=198-180e^{-\frac95\ln\left(\frac53\right)}

Using a calculator,

θ=126.229…\theta=126.229\ldots

Therefore

θ=126∘C\boxed{\theta=126^\circ\text{C}}

to 3 significant figures.

(d)

Differentiate

θ=198−180e−kt\theta=198-180e^{-kt}

with respect to tt:

dθdt=180ke−kt\frac{d\theta}{dt}=180ke^{-kt}

At

t=9t=9

we get

dθdt=180ke−9k\frac{d\theta}{dt}=180ke^{-9k}

Using

k=15ln⁡(53)k=\frac15\ln\left(\frac53\right)

we obtain

dθdt=7.332…\frac{d\theta}{dt}=7.332\ldots

Therefore

7.33 ∘C min−1\boxed{7.33\ ^\circ\text{C min}^{-1}}

to 3 significant figures.