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, theta = 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 Integration: A bottle of water is put into a

A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 15 - 2013 - Paper 1

Question 15

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 15 - 2013 - Paper 1

Step 1

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

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Answer

To solve the differential equation, we start with:

\frac{d\theta}{dt} = \frac{3 - \theta}{125}

We can separate the variables:

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

Integrating both sides:

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

This yields:

-ln(3 - \theta) = \frac{t}{125} + C

Rearranging gives us:

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

Substituting $e^C$ with A, we have:

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

Thus:

\theta = A e^{-\frac{t}{125}} + 3

To express $A$ in terms of an initial condition, when $t = 0$ , $\theta = 16$ :

16 = A e^{0} + 3\ A = 16 - 3 = 13.

Therefore, we arrive at:

$\theta = 13 e^{-0.008t} + 3,$ which concludes the proof.

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

We need to find the time $t$ when $\theta = 10$ . Starting from:

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

Subtracting 3 from both sides gives:

7 = 13 e^{-0.008t}.

Dividing both sides by 13:

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

Taking the natural logarithm of both sides:

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

Thus, we can express $t$ as:

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

Calculating:

substituting in gives:

Rounding to the nearest minute, the time taken for the temperature of the water in the bottle to fall to 10 °C is approximately 25 minutes.

A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 15 - 2013 - Paper 1

Worked Solution & Example Answer:A bottle of water is put into a refrigerator - Edexcel - A-Level Maths Pure - Question 15 - 2013 - Paper 1

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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