Suppose a 95 degree C cup of coffee sits in a 20 degree C room. Let y(t) be the

Question

Suppose a 95 degree C cup of coffee sits in a 20 degree C room. Let y(t) be the temperature of the coffee at time t minutes. Newton's Law of Cooling specifies that dy/dt = k(y - T). where T is the surrounding temperature and k is the proportionality constant. Find the value of k if the coffee cools at a rate of 1 degree C per minute when its temperature is 50 degree C. Solve the differential equation to find an expression for the temperature of the coffee at any time t. What is the temperature of the coffee t = 10 minutes later? (Round your answer to two decimal places.)

Explanation / Answer

(1) Given that the surrounding temperature is 20°C, Newton's Law of Cooling gives the following DE that relates the temperature of the coffee to the time t in minutes that it has been sitting out:
dT/dt = k(T - 20).

We are given that the coffee cools at a rate of 1°C/min when the coffee is 50°C, so dT/dt = -1 (note that this is negative since T is decreasing) when T = 50; this gives us the value of k to be:
-1 = k(50 - 20) = 30k ==> k = -1/30.

(2) Intuitively, the limiting temperature of the coffee will be the temperature of the room. This can be shown by solving the DE and letting t --> infinity.

To solve this DE, we need to separate the variables. Doing this gives:
dT/(T - 20) = -1/30 dt.

Integrating both sides:
ln|T - 20| = (-1/30)t + C, where C is a constant
==> T = 20 + e^[(-1/30)t + C] = 20 + C*e^(-t/30), where C = e^(C).

Given that the initial temperature of the coffee is 95°C, we see that T = 95 when t = 0, so the value of C is:
95 = 20 + C ==> C = 95 - 20 = 75,

giving us the equation:
T = 20 + 75e^(-t/30).

Since 75e^(-t/30) --> 0 as t --> infinity, T --> 20 as t --> infinity, yielding the limiting temperature to be 20°C

(3) Temparature of coffee after 10 minutes ,
dT/dt = (-1/30)(10 - 20) = (-1/30) * (-10) = 1/300 = 0.0033degree C.

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