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 deter

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

0 thoughts on “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 deter”

  1. This is a Newton’s Law of Cooling application. 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 Cooling equation 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 initial coffee temperature .
    T_{S} = 70^{\circ} is the surrounding air temperature .
    k = ? is a constant given by the experiment.
    Compute constant k . The question gives experimental results over one minute’s time.
    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. 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 time t 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 temperature to 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

    Reply

Leave a Comment