Differential Equations
Applications of differential equations
Grade 12
Question:
<p>The rate of cooling of a substance in moving air is proportional to the difference of temperatures of the substance and the air. A substance cools from 36°C to 34°C in 15 minutes. Find when the substance will have the temperature 32°C, it being known that the constant temperature of air is 30°C.</p>
<p>\(t = \frac{15\log 3}{\log \frac{3}{2}}\) minutes</p>
<p>\(t = \frac{15\log 2}{\log \frac{3}{2}}\) minutes</p>
<p>\(t = \frac{15\log 3}{\log 3}\) minutes</p>
<p>\(t = \frac{15\log 2}{\log 3}\) minutes</p>
Step-by-Step Solution
Key Concept: Newton's Law of Cooling states dT/dt = -k(T - T_air), which is a separable differential equation. Solving it gives T = T_air + (T_0 - T_air)e^(-kt), where the exponential decay determines when T reaches 32°C.
<p><strong>Step 1:</strong> Set up Newton's Law of Cooling: dT/dt = -k(T - 30), where T_air = 30°C</p><p><strong>Step 2:</strong> Separate variables and integrate: dT/(T - 30) = -k·dt → ln(T - 30) = -kt + C</p><p><strong>Step 3:</strong> General solution: T - 30 = Ae^(-kt) or T = 30 + Ae^(-kt)</p><p><strong>Step 4:</strong> Apply initial condition T(0) = 36°C: 36 = 30 + A → A = 6, so T = 30 + 6e^(-kt)</p><p><strong>Step 5:</strong> Use T(15) = 34°C to find k: 34 = 30 + 6e^(-15k) → 4 = 6e^(-15k) → e^(-15k) = 2/3 → k = ln(3/2)/15</p><p><strong>Step 6:</strong> Find time when T = 32°C: 32 = 30 + 6e^(-kt) → 2 = 6e^(-kt) → e^(-kt) = 1/3</p><p><strong>Step 7:</strong> -kt = ln(1/3) → t = ln(3)/k = ln(3)/(ln(3/2)/15) = 15ln(3)/ln(3/2) = 15ln(3)/(ln3 - ln2)</p><p><strong>Step 8:</strong> Simplify: t = 15ln(3)/ln(3/2) ≈ 45 minutes (or t = 15[ln(3) - ln(2)]/ln(3/2) × [ln3/ln(3/2)] = 45 minutes)</p><p>∴ Answer: 45 minutes (Option A)</p>
Correct Answer: A