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Thread: Exponential growth and decay

  1. #1
    Senior Member Vinod's Avatar
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    Exponential growth and decay

    Hello Forumites,
    A cup of coffee, cooling off in a room at a temperature $20^{\circ}C,$ has cooling constant $k=0.09min^{-1}$. Now a)I want to know how fast is the coffee cooling(in degrees per minute) when its temperature is $T=80^{\circ}C?$

    b) I want to use linear approximation to estimate the change in temperature over the next 6 seconds when $T=80^{\circ}C$

    c) The coffee is served at a temperature of $90^{\circ}C.$ How long should I wait before drinking it if the optimal temperature is $65^{\circ}C$?

    Answer:-a)$ \frac{dT}{dt}|_{T=80^{\circ}C}=-0.09*(80-20)=-5.4^{\circ}C/min.$

    b) Note that 6 seconds should be used as 0.1 minutes.From $T\approx 80-5.4\Delta t=80-5.4*0.1$. It follows that the change of temperature will be $T-80\approx -0.54^{\circ}C$

    c) How to answer c)? Answer provided to me is 6.4 minutes approx.
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  2. #2
    Senior Member Vinod's Avatar
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    Re: Exponential growth and decay

    Hello,
    As per Newton's law of cooling,$e^{-0.09t}=\frac{65-20}{90-20}$ which comes to 4.91 minutes approx.
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  3. #3
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    Re: Exponential growth and decay

    Which of your questions does that last post answer?

    a) Newton's law of cooling basically says that the rate at which heat flows from a warmer to a cooler body is proportional to the difference in temperatures. When the temperatures are 80 and 20, that difference is 60 degrees Celsius. Given that the constant of proportionality is 0.09 degrees Celsius per minute, the rate at which the coffee is cooling is (0.09)(60)= 5.4 degrees Celsius per minute, as you say.

    b) A "linear approximation" or any linear equation is of the form T= at+ b where T is the temperature at time t. The value at time t= 0 is, of course, b and the rate of change is a. Here, the coffee has T= 80 at t= 0 so b= 805 and, from the previous problem, a= -5.4. The linear approximation is T(t)= 80- 5.4t. In 6 seconds, the temperature will be T(6)= 80- 5.4(0.1)= i80- 0.54= 79.46 degrees Celsius. Hardly noticeable! But, again, that is what you have.

    c) Given that the initial temperature was 90 rather than 80, we have to calculate (a) again. 90- 20= 70 so the rate at which the coffee is cooling is .09(90)= 8.1 degrees Celcius per minute. The temperature, at time t, given that T(0)= 90, will be T(t)= 90- 8.1t. Using that linear approximation, you need to solve the equation T(t)= 90- 8.1t= 65. (This does NOT give 6.4 minutes!)

    However, you titled this "exponential-growth-decay" and using a linear approximation is NOT using "exponential-growth-decay"! Newton's law of cooling, applied here, gives the differential equation, dT/dt= -0.09T. We can write that as dT/T= -0.09 dt. Integrating both sides, ln(T)= 0.09t+ C so that T(t)= e^{-0.09t+ C}= C'e^{-0.09t} where C'= e^C.

    That is, the true temperature function, of which the above is a "linear approximation", is T(t)= C'e^{-0.09t}. If the initial temperature of the coffee is 80 degrees Celsius then C'e^{-0.09(0)}= C'= 80. T(t)= 80 e^{-0.09t}. In 6 seconds= 0.1 minute, the temperature of the coffee will be T(0.1)= 80 e^{-0.009}= 80(0.991)= 79.28 degrees Celsius, not 79.46.

    For (c) you want to solve T(t)= 90 e^{-0.09t}= 65.
    Thanks from Vinod and topsquark
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    Senior Member Vinod's Avatar
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    Re: Exponential growth and decay

    Hello HallsofIvy,
    Newton's law of cooling states that "temperature at certain time=surrounding teperature +(starting temperature - surrounding temperature)* exponential to the power of (-cooling constant *time)"
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    Re: Exponential growth and decay

    I would consider that a result of "Newton's Law of Cooling" rather than the law itself. Of course, you need to solve a differential equation to go from my version to yours so if you have not yet studied differential equations, you might well have been given that version to use.
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