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

The objective is to determine the temperature \( T(45) \) using the given differential equation \( \frac{dT}{dt} = -K(T - 80) \), with the initial condition \( T(0) = 160^\circ F \) and the condition \( T(15) = 120^\circ F \).

1. First, solve the differential equation. This is a first-order linear differential equation. Separate the variables:

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

1. Integrate both sides:

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

1. The integration yields:

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

1. To solve for \( T \), exponentiate both sides:

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

1. This can be rewritten as:

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

1. Therefore, \( T(t) \) is 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. The equation is now:

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

1. 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 gives:

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

1. Substitute these values to find \( T(45) \):

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

This simplifies to:

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

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

The correct answer is \(90^\circ F\).