Hint
State the ambient value and sign of the rate constant. Subtract L from T(0).
Worked solution
(a)
Separate \(\displaystyle \frac{dT}{T- L}= - kdt\) to obtain ln|T−L|=−kt+C. The initial positive excess A gives \(\displaystyle T-L=Ae^{−kt}\).
Answer: \(\displaystyle \frac{dT}{dt}=-\frac{1}{10}(T-17)\); \(\displaystyle T=17 + 10 e^{- \frac{t}{10}}\).
(b)
Set \(\displaystyle T- L= \frac{A}{2}\), cancel the positive A, and solve \(\displaystyle e^{− kt}= \frac{1}{2}\). For arbitrarily large t, the exponential tends to zero from above.
Answer: \(10 \log{\left(2 \right)}\) minutes; T tends to 17°C.
(c)
The exponential excess tends to zero.
\(17\)
Answer: \(17\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 3 marks
- 1 mark: Translate proportional cooling into the correctly signed separable equation.
- 1 mark: Separate, integrate and apply the initial temperature.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Cooling uses a negative coefficient.
(b) · 2 marks
- 1 mark: Halve the excess above ambient and solve for time, then use exponential decay to obtain the limit.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Half of the excess is not half of the absolute temperature.
(c) · 2 marks
- 1 mark: The exponential excess tends to zero.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Cooling uses a negative coefficient.
Total: 7 marks.