Water is being heated in an electric kettle - Edexcel - A-Level Maths Pure - Question 6 - 2015 - Paper 3

Given that θ = 70°C when t = 40, substitute these values into the equation:

$$70 = 120 - 100 e^{-\lambda \cdot 40}.$$

Rearranging this gives:

$$100 e^{-40\lambda} = 120 - 70 = 50.$$

Dividing both sides by 100:

$$e^{-40\lambda} = 0.5.$$

Taking the natural logarithm on both sides results in:

$$-40\lambda = \ln(0.5).$$

Thus,

$$\lambda = -\frac{\ln(0.5)}{40} = \frac{\ln(2)}{40}.$$

So, the exact value of λ is ( \frac{\ln 2}{40} ).

To find T when θ = 100°C, substitute θ into the equation:

$$100 = 120 - 100 e^{-\lambda T}.$$

Rearranging gives:

$$e^{-\lambda T} = \frac{120 - 100}{100} = 0.2.$$

Taking the natural logarithm:

$$-\lambda T = \ln(0.2).$$

Substituting λ from part (b):

$$-\frac{\ln(2)}{40} T = \ln(0.2).$$

This can be rewritten as:

$$T = -\frac{40 \ln(0.2)}{\ln(2)}.$$

Calculating:

$$\ln(0.2) = \ln\left(\frac{2}{10}\right) = \ln(2) - \ln(10),\ \ln(10) \approx 2.302.$$

This gives approximately:

$$T = -\frac{40 (\ln(2) - 2.302)}{\ln(2)} ∼ 93.5.$$

Rounding to the nearest whole number, we find T = 93.