T= temperature of coffee at time t

t = time (any units, but to be consistent with the given information, choose minutes)

T_{a} = temperature of the surrounding atmosphere = 50 degrees

T_{0} = initial temperature of coffee = 200 degrees

The temperature is proportional to the difference between its current temperature and it surrounding temperature. We can express this mathematically using this differential equation

dT/dt = -k(T-T_{a})

Separate the variables .

dT/(T-T_{a}) = -kdt

Integrate.

ln (T-T_{a}) = -kt +C

Eliminate the natural log.

(T-T_{a}) = e^{-kt+C}

(T-T_{a}) = Ce^{-kt}

When t = 0, what is C? It's simply the initial difference in temperatures. C = (T_{0} - T_{a})

(T-T_{a}) = (T_{0} - T_{a})e^{-kt}

and the problem asks you to represent T-T_{a }with C(t).

C(t) = (T_{0} - T_{a})e^{-kt
}For this specific problem, the rate of cooling is 2 degrees/min at t=0, which means

dT/dt = -k(T-T_{a})

-2 deg/min = -k (200-50)

k = 2/150 = 1/75 ~ 0.01333

and of course T_{0} - T_{a } = 200 deg - 50 deg = 150 deg

Substituting into our expression for C(t), we finally have the expression for the difference of temperature of the coffee and the surrounding atmosphere as a function of time

C(t) = (150 deg)e^{-0.01333t}

If the milk cools to 120 degrees, then C(t) = 120 deg - 50 deg = 70 deg.

70 deg = (150 deg)e^{-0.01333t}

Solve for t.

70/150 = e^{-0.01333t}

ln (70/150) = -0.01333t

t = ln(70/150)/(-0.01333)

t = -0.7621/(-0.01333)

**t = 57.16 min**

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