Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
If the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of air is 30°C and the substance cools from 37°C to 34°C in 15 min then:
Temperature of substance will be 32°C at $t = 15\log_{7/4}\frac{2}{7}$ min
Temperature of substance will be 31°C at $t = 15\log_{7/4}7$ min
Proportional constant $k = \frac{1}{15}\log\frac{7}{4}$
All of these
Step-by-Step Solution
Key Concept: Newton's Law of Cooling states dθ/dt = -k(θ - θ₀). Using the boundary condition that temperature drops from 37°C to 34°C in 15 minutes with ambient temperature 30°C, we solve the separable differential equation to find k = (1/15)ln(7/4), then use this to calculate times for any target temperature.
Newton's cooling law gives $\frac{d\theta}{dt} = -k(\theta - \theta_0)$ with $\theta_0 = 30°C$. Solving yields $\ln|\theta - 30| = -kx + C$. Using boundary conditions to find $k = \frac{1}{15}\ln\frac{7}{4}$, the time to reach $31°C$ is $t(31°) = 15\log_{7/4}7$ and to reach $32°C$ is $t(32°) = 15\log_{7/4}\frac{7}{2}$.
Correct Answer: 1,2,3,4