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A quantity of ethanol was heated until it reached boiling point - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 2

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A quantity of ethanol was heated until it reached boiling point. The temperature of the ethanol, θ °C, at time t seconds after heating began, is modelled by the equ... show full transcript

Worked Solution & Example Answer:A quantity of ethanol was heated until it reached boiling point - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 2

Step 1

Find A and B

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Answer

To find the values of A and B:

  1. At t = 0, the temperature θ = 18 °C, so substituting into the equation gives:

    18 = A - B$$ This simplifies to: $$A - B = 18 \, (1)$$
  2. At t = 10 seconds, θ = 44 °C, substituting this value into the equation gives:

    44 = A - Be^{-0.7}$$ Rearranging gives: $$A - 0.4966B = 44 \, (2)$$ (using the value of e^{-0.7} ≈ 0.4966)
  3. Now we can solve the equations (1) and (2) together:

    B=18AB = 18 - A

    Substitute into (2):

    A - 8.92 + 0.4966A = 44 \ 1.4966A = 52.92 \ A = 35.39$$ 4. Now substitute A back to find B: $$B = 18 - 35.39 = -17.39$$

Thus, the complete model with A and B is:

θ=35.39(17.39)e0.07tθ = 35.39 - (-17.39)e^{-0.07t}

or approximately, rounding to three significant figures:

θ=35.4+17.4e0.07tθ = 35.4 + 17.4e^{-0.07t}

Step 2

Evaluate the model

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Answer

To evaluate the model:

  1. The boiling point of ethanol is approximately 78 °C.

  2. The maximum temperature predicted by the model occurs as t approaches infinity and can be determined from:

    extlimtoext(35.4+17.4e0.07t)=35.4+0=35.4°C ext{lim}_{t o ext{∞}} (35.4 + 17.4 e^{-0.07t}) = 35.4 + 0 = 35.4 °C

  3. Since 35.4 °C is much lower than 78 °C, the model is not appropriate for real-world conditions, as it does not reach the boiling point. Hence, this indicates that the model fails to accurately represent the behavior of ethanol when heated.

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