Using the differential equation modeling Newton's Law of Cooling dTdt=k(T−Te)dTdt=k(T−Te), Answer the following. Brewing Coffee: The brewing temperature of the water used is very important. It should be between 195 F and 205 F. The closer to 205 F the better. Boiling water (212 F) should never be used, as it will burn the coffee. Water that is less than 195 F will not extract properly. On the other hand, coffee that has a temperature of 205 F is too hot to drink. Coffee is best when it is served at a temperature of 140 F to 155 F (the Goldilocks range). Suppose coffee is initially brewed at 205 F and the room temperature is 70 F. Determine the value of kk if the temperature of the coffee drops from 205 F to 200 F in the first two minutes after brewing. Round answer to 4 decimal places.

   k = -3.1450 10⁻⁴  s⁻¹

Explanation:

In this exercise we are given the equation that describes the cooling process

        dT / dt = k (T -)

Let's solve is this equation,

        dT / (T-T_ {e}) = k dt

change of variable for integration

       T -T_{e} = T ’

       dT = dT '

       ∫ dT ’/ T’ = k  ∫ dt

we integrate

        ln T ’= k t

we change to the initial variables

        ln (T - T_{e}) = k t

Let's evaluate from the lower limit T = T for t = 0 to the upper limit T = T₀ for time t

       ln (T₀ -T_{e}) - ln (T -T_{e}) = k (t-0)

we simplify

       ln (T₀ -T_{e} / T -T_{e}) = k t

       k = ln (T₀ -T_{e})  / (T-Te) / t

In the exercise they indicate that the temperature T = 205 F, the ambient temperature is T_{e} = 70F, the temperature to which T₀ = 200 F falls in a time t = 2 min = 120 s

Let's calculate

           k = ln [(200- 70) / (205 -70)] / 120

           k = -0.0377403 / 120

           k = -3.1450 10⁻⁴  s⁻¹

mistakes are ok

see earlier answer