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

The Correct Option is C

To solve this problem, we need to find the temperature \( T(45) \) given the differential equation \( \frac{dT}{dt} = -K(T - 80) \), with initial condition \( T(0) = 160^\circ F \) and an additional piece of information \( T(15) = 120^\circ F \).

1. Start by solving the differential equation. We recognize this as a first-order linear differential equation. To solve it, let's separate the variables:

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

1. Integrate both sides:

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

1. The integrals result in:

\(\ln|T - 80| = -Kt + C\)

1. Solving for \( T \), we exponentiate both sides:

\(|T - 80| = e^{-Kt + C} = Ce^{-Kt}\)

1. We can write this as:

\(T - 80 = C'e^{-Kt}\) (considering \( C' = \pm C \))

1. Thus, \( T(t) \) can be expressed as:

\(T(t) = 80 + C'e^{-Kt}\)

1. Use the initial condition \( T(0) = 160 \) to find \( C' \):

\(160 = 80 + C'e^0 \Rightarrow C' = 80\)

1. Thus, the equation becomes:

\(T(t) = 80 + 80e^{-Kt}\)

1. Now, use the condition \( T(15) = 120 \) to find \( K \):

\(120 = 80 + 80e^{-15K} \Rightarrow 40 = 80e^{-15K} \Rightarrow e^{-15K} = \frac{1}{2}\)

1. Taking the natural logarithm, we find:

\(-15K = \ln\left(\frac{1}{2}\right) \Rightarrow K = \frac{\ln 2}{15}\)

1. Substitute back to find \( T(45) \):

\(T(45) = 80 + 80e^{-45\cdot\frac{\ln 2}{15}} = 80 + 80e^{-3 \ln 2} = 80 + 80e^{-\ln 8}\)

Simplifies to:

\(T(45) = 80 + 80\cdot\frac{1}{8} = 80 + 10 = 90\)

1. Thus, the temperature \( T(45) \) is \( 90^\circ F \).

The correct answer is \(90^\circ F\).

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