Question 3(e) Merit to Excellence

Two exponential terms to integrate, a log-form differential equation, an upper limit that appears on both sides, a bisected cosine area, and Newton’s law of cooling built from scratch.

A teacher makes a cup of coffee at the start of interval and leaves it in the staff room, where the temperature is 18 °C. After 30 minutes the coffee is 50 °C. After a further one hour it has cooled to 30 °C. The rate at which the temperature changes at any instant is proportional to the difference between the temperature of the coffee, N, and the temperature of the room.

Write a differential equation that models this situation, and then solve it to calculate the temperature of the cup of coffee when it was made.

The 2024 paper, page 13 ↗ Marking schedule, page 7 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Turn the sentence into a differential equation. The rate of change of N is proportional to what?

   To the difference between N and the room’s 18 °C:
   dN/dt = k(N − 18)

2. 2 Separate and integrate.

   int (1/(N − 18)) dN = int k dt
   ln |N − 18| = kt + c

3. 3 Measure t in minutes from when the coffee was made. Which two facts do you have?

   t = 30, N = 50: ln 32 = 30k + c
   t = 90, N = 30: ln 12 = 90k + c

4. 4 Subtract one from the other to find k, then find c.

   60k = ln 12 − ln 32 = ln3/8, so k = (ln(3/8))/60 ≈ −0.01635.
   c = ln 32 − 30k ≈ 3.95615

5. 5 The coffee was made at t = 0. What is N then?

   ln |N − 18| = c, so N − 18 = e^3.95615 ≈ 52.26 and N ≈ 70.26 °C.
   Exactly, N − 18 = 32√(8/3).

What the marker wanted

• Achieved Reaching ln |N − 18| = kt + c.
• Merit Both k and c found, with correct integration. Any sensible rounding is accepted.
• Excellence The starting temperature, 70.26 °C, from correct integration, is E8; a solution with a minor error is E7. Working in hours gives k ≈ −0.9808 and the same 70.26 °C.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

Got it moves it further out. Review brings it back tomorrow.