Exponential growth and decay practice question 8
A kettle of water at 100 °C is left to cool in a room of unknown constant temperature. After 10 minutes the water is at 80 °C, and after 20 minutes it is at 65 °C. Assuming Newton’s law of cooling, find the room temperature.
Give the answer in degrees Celsius.
There are two unknowns, the room temperature and the decay constant, and the readings are 10 minutes apart. What does each 10-minute step do to the gap between water and room?
Let the room be . Then , with two unknowns and , and two readings.
The readings are equally spaced, so the ratio equals the ratio ; both equal . That equation has no in it.
A number.
Theory UBC Math 100 §11.4.1CLP-1 §3.3Key results
Worked solution
Strategy
With two unknowns you need two readings, but the decay constant can be eliminated: over equal time steps the temperature difference is multiplied by the same factor. Setting the two factors equal gives an equation in the room temperature alone.
Working
Newton’s law with unknown surroundings A.
Each 10-minute step multiplies the difference by e10k.
Cross-multiply and expand; the A² terms cancel.
The differences 80, 60, 45 each fall by a factor of 3/4: consistent.
Answer
The answer is 20 degrees Celsius.
Check
With a 20 °C room the gaps 80, 60 and 45 fall by the same factor 3/4 every 10 minutes, as exponential decay requires.
Where this goes wrong
Assuming a room temperature, such as 0 °C, and using only one reading. The problem gives two readings precisely because the room temperature is unknown and has to be solved for.
The idea behind it
The exponential growth and decay lesson, a figure to move, in a few steps
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