A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at 3 °C and t minutes after the bottle is placed in the refrigerator the temperature of the water in the bottle is θ °C. The rate of change of the temperature of the water in the bottle is modelled by the differential equation, dθ/dt = (3 - θ) / 125 (a) By solving the differential equation, show, that, θ = Ae^{-0.008t} + 3 where A is a constant. Given that the temperature of the water in the bottle when it was put in the refrigerator was 16 °C, (b) find the time taken for the temperature of the water in the bottle to fall to 10 °C, giving your answer to the nearest minute.

A-Level Edexcel Maths Pure Differentiation: A bottle of water is put into a

A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 7

Question 2

A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at 3 °C and t minutes after the bottle is placed in the refrig... show full transcript

Worked Solution & Example Answer:A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 7

Step 1

By solving the differential equation, show, that, θ = Ae^{-0.008t} + 3

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Answer

To solve the differential equation ( \frac{d\theta}{dt} = \frac{(3 - \theta)}{125} ), we separate variables:

$\int \frac{1}{3 - \theta} d\theta = \int \frac{1}{125} dt$

This gives:

$-\ln|3 - \theta| = \frac{t}{125} + c$

Proceed by exponentiating both sides to eliminate the logarithm:

$3 - \theta = e^{-\frac{t}{125} - c}$

Letting ( A = e^{-c} ) allows us to rewrite this as:

$3 - \theta = Ae^{-\frac{t}{125}}$

Rearranging results in:

$\theta = 3 - Ae^{-\frac{t}{125}}$

To fit the desired form, we replace ( A ) with ( -A ) and re-arrange, yielding:

$\theta = Ae^{-0.008t} + 3$

where ( A ) is a constant that depends on the initial conditions.

Step 2

find the time taken for the temperature of the water in the bottle to fall to 10 °C

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Answer

From the initial condition, we know when ( t = 0 ), ( \theta = 16 ):

$16 = Ae^{0} + 3 \Rightarrow A = 13$

Substituting ( A ) into the equation:

$\theta = 13e^{-0.008t} + 3$

Now set ( \theta = 10 ) to find the time ( t ):

$10 = 13e^{-0.008t} + 3$

This simplifies to:

$7 = 13e^{-0.008t}$

Further simplifying gives:

$e^{-0.008t} = \frac{7}{13}$

Taking the natural logarithm of both sides:

$-0.008t = \ln\left(\frac{7}{13}\right)$

Now solving for ( t ):

$t = -\frac{\ln\left(\frac{7}{13}\right)}{0.008}$

Calculating the value:

$t \approx 77.3799 \text{ minutes}$

Thus, rounding to the nearest minute, the time taken is approximately 77 minutes.

A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 7

Worked Solution & Example Answer:A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 7

By solving the differential equation, show, that, θ = Ae^{-0.008t} + 3

find the time taken for the temperature of the water in the bottle to fall to 10 °C

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