The temperature of a cup of coffee varies according to Newton’s Law of Cooling: -“dT/dt=k(T-A), where is the temperature of the coffee, A is

The temperature of a cup of coffee varies according to Newton’s Law of Cooling: -“dT/dt=k(T-A), where is the temperature of the coffee, A is the room temperature, and k is a positive
constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25*C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes,
74
67
60
42

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  1. Answer:

    B) 67°C.

    Step-by-step explanation:

    Newton’s Law of Cooling is given by:

    [tex]\displaystyle \frac{dT}{dt}=k(T-A)[/tex]

    Where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.

    We are given that the coffee cools from 100°C to 90°C in one minute at a room temperature A of 25°C.

    And we want to find the temperature of the coffee after four minutes.

    First, solve the differential equation. Multiply both sides by dt and divide both sides by (TA). Hence:

    [tex]\displaystyle \frac{dT}{T-A}=k\, dt[/tex]

    Take the integral of both sides:

    [tex]\displaystyle \int \frac{dT}{T-A}=\int k\, dt[/tex]

    Integrate:

    [tex]\displaystyle \ln\left|T-A\right| = kt+C[/tex]

    Raise both sides to e:

    [tex]|T-A|=e^{kt+C}=Ce^{kt}[/tex]

    The temperature of the coffee T will always be greater than or equal to the room temperature A. Thus, we can remove the absolute value:

    [tex]\displaystyle T=Ce^{kt}+A[/tex]

    We are given that A = 25. Hence:

    [tex]\displaystyle T=Ce^{kt}+25[/tex]

    Since the coffee cools from 100°C to 90°C, the initial temperature of the coffee was 100°C. Thus, when t = 0,T = 100:

    [tex]100=Ce^{k(0)}+25\Rightarrow C=75[/tex]

    Hence:

    [tex]T=75e^{kt}+25[/tex]

    We are given that the coffee cools from 100°C to 90°C after one minute at a room temperature of 25°C.

    So, T = 90 given that t = 1. Substitute:

    [tex]90=75e^{k(1)}+25[/tex]

    Solve for k:

    [tex]\displaystyle e^k=\frac{13}{15}\Rightarrow k=\ln\left(\frac{13}{15}\right)[/tex]

    Therefore:

    [tex]\displaystyle T=75e^{\ln({}^{13}\! /\!{}_{15})t}+25[/tex]

    Then after four minutes, the temperature of the coffee will be:

    [tex]\displaystyle \begin{aligned} \displaystyle T&=75e^{\ln({}^{13}\! /\!{}_{15})(4)}+25\\\\&\approx 67^\circ\text{C}\end{aligned}[/tex]

    Hence, our answer is B.

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