Question

The temperature $T(t)$ of a body at time $t = 0$ is $160^\circ$F and it decreases continuously as per the differential equation $\frac{dT}{dt} = -K(T - 80),$ **where $K$ is a positive constant. If $T(15) = 120^\circ$F, then $T(45)$ is equal to

• A. $85\degree F$
• B. $95\degree F$
• C. $90\degree F$
• D. $80\degree F$

Answer 1

Answer: (C)

$90\degree F$

Answer 2

Given:

$\frac{dT}{dt} = -K(T - 80)$

Separate variables and integrate:

$\int_{160}^{T} \frac{1}{T - 80} dT = - \int_{0}^{t} K \, dt$

This gives:

$\left[ \ln |T - 80| \right]_{160}^{T} = -Kt$ $\ln |T - 80| - \ln 80 = -Kt$ $\ln \frac{T - 80}{80} = -Kt$

Exponentiate both sides:

$\frac{T - 80}{80} = e^{-Kt}$ $T = 80 + 80e^{-Kt}$

Use the initial condition $T(15) = 120$ to find $K$

$120 = 80 + 80e^{-K \cdot 15}$ $40 = 80e^{-15K}$ $\frac{1}{2} = e^{-15K}$

Take the natural logarithm:

$-15K = \ln \frac{1}{2} = -\ln 2$ $K = \frac{\ln 2}{15}$

Find $T(45)$. Substitute $t = 45$:

$T(45) = 80 + 80e^{-K \cdot 45}$ $= 80 + 80 \left( e^{-15K} \right)^3$ $= 80 + 80 \left( \frac{1}{2} \right)^3$ $= 80 + 80 \cdot \frac{1}{8}$ $= 80 + 10 = 90$

Thus, the answer is:

90°F