QCE Specialist Maths: Differential equations: piecewise cooling Q160

Q160 · Practice questionTechnology-activeComplex unfamiliar7 marks

QUESTION 160 (7 marks)

A metal block cools according to

\frac{dT}{dt}=-k(T-A),

where

T

and the surrounding temperature

A

are in degrees Celsius,

t

is in minutes and

k>0

is constant. Initially

T(0)=100

and

A=20

. At

t=10

T=60

. At that instant the block is moved to a room at

A=10

. The block's temperature is continuous at the move and

k

is unchanged. The diagram shows the surrounding temperature, not the block temperature.

Ambient temperature A=20 from t=0 to just before 10 minutes, then A=10 from t=10 onwards. An open dot at (10,20) and filled dot at (10,10) distinguish the intervals.

Determine the exact value of

k

, showing how the differential equation is solved.

[3 marks]

Derive

T(t)

for

t\ge10

and determine when the block first reaches

30

degrees Celsius. Give the time since the start to two decimal places.

[4 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

7 marks

ANSWER

(a)

k=\ln2/10

. (b)

T(t)=10+50e^{-k(t-10)}

for

t\ge10

; first reaches

30

degrees Celsius at

t=10+10\ln(5/2)/\ln2\approx23.22

min.

Worked solution

(a) Separation and integration for the first room give

\ln(T-20)=-kt+C

, hence

T=20+80e^{-kt}

. Since

T(10)=60

e^{-10k}=1/2

and

k=\ln2/10

per minute.

(b) Reset elapsed time at the move to

u=t-10

. In the second room, separation gives

T-10=Ce^{-ku}

. Continuity gives

T(10)=60

, so

C=50

and

T(t)=10+50e^{-k(t-10)}.

Set

T=30

e^{-k(t-10)}=2/5

, so

t=10+\frac{\ln(5/2)}{k}=10+\frac{10\ln(5/2)}{\ln2}\approx23.22.

Before the move the block remains at least 60 degrees Celsius; afterwards its derivative is negative while

T>10

, so this is the unique first time. The temperature does not jump when the ambient temperature changes.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Separate and integrate the first-room equation.

[1 mark]

Part a: Use T(0)=100 to obtain T=20+80 exp(-kt).

[1 mark]

Part a: Use T(10)=60 to obtain k=ln(2)/10.

[1 mark]

Part b: Solve the second-room equation with elapsed time t-10.

[1 mark]

Part b: Use continuity to obtain T=10+50 exp(-k(t-10)).

[1 mark]

Part b: Solve for the exact time and 23.22 min.

[1 mark]

Part b: Justify that it is the first time using the two intervals.

[1 mark]

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Differential equations: piecewise cooling — Question 160

QUESTION 160 (7 marks)

Practice marking scheme