Question

Suppose both cups are poured at the same time, and both are left sitting in the room that is 75°F. But this time,

Answer 1

Answer:

The equations of both cups after t minutes is given by:

$T_{Cup1}=75 °F+(180 °F-75 °F)e^{-0.2337min^{-1}*t}$

$T_{Cup1}=75 °F+(162 °F-75 °F)e^{-0.2337min^{-1}*t}$

Step-by-step explanation:

The complete exercise says:

Two cups of coffee are poured from the same pot. The initial temperature of the coffee is 180°F, and is 0.2337 (for time in minutes).

Newton's law of cooling:

$T=T_m+(T_0-T_m)e^{-kt}$

where,

$T_m$= room temperature

$T_0$= initial temperature

k=cooling constant

Hence,

Cup 1:

$T_m$= 75 °F

$T_0$= 180 °F

$T_{Cup1}=75 °F+(180 °F-75 °F)e^{-0.2337min^{-1}*t}$

Cup 2:

$T_m$= 75 °F

$T_0$= 162 °F

$T_{Cup1}=75 °F+(162 °F-75 °F)e^{-0.2337min^{-1}*t}$