Experimenting, you have found that 12 ounces of 180°f coffee in your favorite cup will take 20 minutes to cool to a drinking temperature of 110°f in a 70°f room. assume that when you add cream to the coffee, the two liquids are mixed instantly, and the temperature of the mixture instantly becomes the weighted average of the temperature of the coffee and of the cream (weighted by the number of ounces of each fluid). also assume that the cooling constant of the liquid (the constant k in the equations on p. 237 in section 3.8 for newton's law of cooling) does not change when you add the cream. (a) if you add 2 ounces of 40°f cream to the 180°f coffee in the cup, how long will it take for the mixture to reach drinking temperature? give your answer (in minutes) in decimal form with at least 3 decimal digits (for example, 12.345 minutes).

16.033 min

Step-by-step explanation:

We can write Newton's Law of Cooling as

T = Tₐ + (T₀ - Tₐ)e^(-kt), where

Tₐ = ambient temperature of the surroundings

T₀ = initial temperature of the object

k = the cooling constant

t = the time

(1) Calculate the cooling constant

Data:

T =   110 °F

T₀ = 180 °F

Tₐ =  70 °F

 t =   20 min

Calculation:

110 = 70 + (180 - 70)e^(-k×20)

40 = 110e^(-20k)

0.363 64 = e^(-20k)

-1.0116 = -20k

k = 0.050 580 min⁻¹

(2)After adding cream

(a) Calculate T₀  

2 oz × 40 °F + 12 oz × 180 °F = 14 oz × T₀

80 °F + 2160°F = 14T₀

2240 °F = 14T₀

T₀ = 160 °F

(b) Calculate the time to cool to drinking temperature

110 = 70 + (160 - 70)e^(-0.050 580t)

40 = 90e^(-0.050 580t)

0.444 44 = e^(-0.050 580t)

-0.810 93 = -0.050580t

t = 16.033 min

It will take 16.033 min for the mixture to reach drinking temperature

is only a line? i think so