Answer:
[tex]t \approx 17.690\,min[/tex]
Step-by-step explanation:
This differential equation is a first order linear differential equation with separable variables, whose solution is found as follows:
[tex]\frac{dy}{dt} = - \frac{1}{53} \cdot (y - 17)[/tex]
[tex]\frac{dy}{y-17} = -\frac{1}{53} \, dt[/tex]
[tex]\int\limits^{y}_{y_{o}} {\frac{dy}{y-17} } = -\frac{1}{53} \int\limits^{t}_{0}\, dx[/tex]
[tex]\ln \left |\frac{y-17}{y_{o}-17} \right | = -\frac{1}{53} \cdot t[/tex]
[tex]\frac{y-17}{y_{o}-17} = e^{-\frac{1}{53}\cdot t }[/tex]
[tex]y = 17 + (y_{o} - 17) \cdot e^{-\frac{1}{53}\cdot t }[/tex]
The solution of the differential equation is:
[tex]y = 17 + 74\cdot e^{-\frac{1}{53}\cdot t }[/tex]
Where:
[tex]y[/tex] - Temperature, measured in °C.
[tex]t[/tex] - Time, measured in minutes.
The time when the cup of coffee has the temperature of 70 °C is:
[tex]70 = 17 + 74 \cdot e^{-\frac{1}{53}\cdot t }[/tex]
[tex]53 = 74 \cdot e^{-\frac{1}{53}\cdot t }[/tex]
[tex]\frac{53}{74} = e^{-\frac{1}{53}\cdot t }[/tex]
[tex]\ln \frac{53}{74} = -\frac{1}{53}\cdot t[/tex]
[tex]t = - 53\cdot \ln \frac{53}{74}[/tex]
[tex]t \approx 17.690\,min[/tex]
