The differential equation below models the temperature of a 85 degree C cup of coffee in a 25 degree C room, where it is know that the coffe

Question

The differential equation below models the temperature of a 85 degree C cup of coffee in a 25 degree C room, where it is know that the coffee cools at a rate 1 degree C per minute when its temperature is 75 degree C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in degree C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 85 degree C.) dy/dt = -1/50(y-25)

in progress 0
1 week 2021-09-10T22:48:28+00:00 1 Answer 0

Answers ( )

    0
    2021-09-10T22:49:29+00:00

    Answer:

    y(t)=25+Ce^{-\frac{t}{150}}

    Step-by-step explanation:

    We are given that differential equation

    \frac{dy}{dt}=-\frac{1}{50}(y-25)

    We have to find the expression for the temperature of the coffee at time t.

    Let y be the temperature of the coffee in degree C and t be the time in minutes.

    At t=0 , y=85 degree Celsius

    \frac{dy}{y-25}=-\frac{1}{50}dt

    Integrating on both sides

    \int \frac{dy}{y-25}=-\frac{1}{50}\int dt

    ln(y-25)=-\frac{t}{50}+lnC

    ln (y-25)-ln C=-\frac{t}{50}

    ln(m)-ln (n)=ln\frac{m}{n}

    ln\frac{y-25}{C}=-\frac{t}{50}

    \frac{y-25}{C}=e^{-\frac{t}{50}}

    lnx=y\implies x=e^y

    y-25=Ce^{-\frac{t}{150}}

    Substitute the value t=0 and y=85 then we get

    85-25=Ce^{0}

    60=C

    Substitute the value of C

    Then , we get

    y-25=60e^{-\frac{t}{150}}

    y(t)=25+Ce^{-\frac{t}{150}}

    This is required expression for the temperature of the coffee at time t.

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )