The rate of change of the temperature, $T^{ ext{°C}}$ of a mug of coffee is given by
dT
dt = \frac{1}{25}T - k, \quad 0 \leq t \leq 50
- $t$ is the elapsed time, in minutes, after the coffee is poured into the mug
- $k$ is a constant
- initially, the temperature of the coffee is 100°C
- 10 minutes later the temperature has fallen to 82°C - Scottish Highers Maths - Question 15 - 2015
Question 15
The rate of change of the temperature, $T^{ ext{°C}}$ of a mug of coffee is given by
dT
dt = \frac{1}{25}T - k, \quad 0 \leq t \leq 50
- $t$ is the elapsed time, i... show full transcript
Worked Solution & Example Answer:The rate of change of the temperature, $T^{ ext{°C}}$ of a mug of coffee is given by
dT
dt = \frac{1}{25}T - k, \quad 0 \leq t \leq 50
- $t$ is the elapsed time, in minutes, after the coffee is poured into the mug
- $k$ is a constant
- initially, the temperature of the coffee is 100°C
- 10 minutes later the temperature has fallen to 82°C - Scottish Highers Maths - Question 15 - 2015
Step 1
Integrate the term
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Answer
We start with the given differential equation:
dT = \left( \frac{1}{25}T - k \right) dt$$
Integrating both sides,
$$\int dT = \int \left( \frac{1}{25}T - k \right) dt$$
Step 2
Complete integration
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Answer
Integrating the left side gives:
T=251Tt−kt+C
Rearranging results in:
T−251Tt=−kt+C
Step 3
Find constant of integration
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Answer
To find the constant C, we know that initially, at t=0, T=100°C:
100=0−k(0)+C⟹C=100
Step 4
Find value of $k$
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Answer
After 10 minutes, the temperature falls to 82°C, so we set up the equation:
82=251(82)⋅10−10k+100
This simplifies to:
−10k=82−32.8−100⟹k=1051.8=5.18
Step 5
State expression for $T$
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Answer
Substituting k back into the expression for T, we get:
T=251(Tt)−(5.18)t+100
This can also be simplified to:
T(t)=25100−5.18tt
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