If the coffee has a temperature of 200°F when freshly poured and 1 minute later has cooled to 190°F in a room at 70°F, how do you determine when the coffee reaches a temperature Of 150°F?

You can check the answer of the people under the question at Quora “you place a cup of 210 coffee on a table“

This is a Newton’s Law of Coolingapplication. The coffee is the substance being cooled and the atmosphere is the cooling medium.Newton’s Law of Cooling Reference

Here is the Newton’s Law of Coolingequation for temperature T(t) over time. This is the solution to a differential equation but Calculus isn’t needed to solve this.T(t) = T_{S} + (T_{0} − T_{S})e^{−kt}

T_{0} = 200^{\circ} is the initialcoffee temperature .T_{S} = 70^{\circ} is the surrounding air temperature.k = ? is a constantgiven by the experiment.Compute constant k . The question gives experimental resultsover one minute’s time.The coffeehas a temperature of 200°F when freshly poured and 1 minute later has cooled to 190°F in a room at 70°F. Enter these values into the equation and solve for k .T(t) = T_{S} + (T_{0} − T_{S})e^{−kt}

T(1) = 190 = 70 + (200 − 70)e^{k \times (-1)}

Recall that (-1)\ln(x) = \ln(\frac{1}{x}) for the next line.

120 = 130e^{−k} \iff k = (-1)\ln(\frac{12}{13}) \iff k = \ln(\frac{13}{12})

Compute the timet when the coffee reaches the temperature 150°F.T(t) = T_{S} + (T_{0} − T_{S})e^{−kt}

T(t) = 70 + 130e^{−t\ln(\frac{13}{12})} is the general function.Set the temperatureto 150°F and solve for t .150 = 70 + 130e^{−t\ln(\frac{13}{12})}

t = \frac{\ln(\frac{13}{8})}{\ln(\frac{13}{12})} \approx 6.067 minutes

Apply Newton’s Law of Cooling.