## We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the carton, T, as

Question

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the carton, T, as a function of time follows Newton’s cooling law:
dT / dt = k (T – a),
where a is the temperature of the surroundings and k is a constant. We assume the temperature
the surroundings are a = 20 (given in ◦C), and that T (0) = 4 (given in ◦C).
a) Show that T (t) = 20 – 16e ^ kt (given in ◦C).
b) After 5 minutes the temperature of the carton is 8◦C. What is the temperature after 15 minutes?

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1. We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation We now separate the time and the temperature variables as follows, Integrating both sides of the differential equation, we obtain; This natural logarithmic equation can be rewritten as; Applying the laws of exponents, we obtain,  We were given the initial conditions, Let us apply this condition to obtain;  Now our equation, becomes or When we substitute a=20,
we obtain, b) We were also given that, Let us apply this condition again to find k. This implied   We take logarithm to base e of both sides, This implies that,  After 15 minutes, the temperature will be,   After 15 minutes, the temperature is approximately 20°C