### Hot Coffee, give it a few minutes

Today in class we wrapped the last part of the course, except for the coroner problem which we will be doing next class. We did a problem that is similar to the coroner problem, the problem we did a problem about coffee cooling.

T' = k(T-70)

dT/dt = k (T-70)

(1/(T-70))dT = k dt

Now antidifferentiate both sides of the equation.

ln(T-70) = kt+C

T-70 = e^(kt+C)

Now that the variables are separated, use the point (0,190) (T,t) to solve for one of the unknown variables.

T-70 = e^(kt)*e^C

190-70 = e^(k(0))*e^C

120 = 1*e^C

120 = e^C

T-70 = 120e^(kt)

Now that there is only one one unknown "k", use a second point (5,180) to solve the last unknown.

180-70 = 120e^5k

110 = 120e^(5k)

110/120 = e^(5k)

11/12 = e^(5k)

ln(11/12) = 5k

(1/5)ln(11/12) = k

k = -0.0174

T-70 = 120e^(-0.0174t)

Now that the equation has no unknowns besides T, and t, the problem can be answered. Use T=120

120-70 = 120e^(-0.0174t)

50/120 = e^(-0.0174t)

5/12 = e^(-0.0174t)

ln(5/12) = -0.0174t

(1/-0.0174)ln(5/12) = t

t = 50.3143

The coffee is drinkable after 50 minutes.

The next scribe is Sarah

*The coffee's temperature is 190 degrees, and room temperature is 70 degrees, and coffee is "drinkable" at 120 degrees*.*First thing you have to do is identify the differential equation that describes the cooling of the coffee.*T' = k(T-70)

dT/dt = k (T-70)

(1/(T-70))dT = k dt

Now antidifferentiate both sides of the equation.

ln(T-70) = kt+C

T-70 = e^(kt+C)

Now that the variables are separated, use the point (0,190) (T,t) to solve for one of the unknown variables.

T-70 = e^(kt)*e^C

190-70 = e^(k(0))*e^C

120 = 1*e^C

120 = e^C

T-70 = 120e^(kt)

Now that there is only one one unknown "k", use a second point (5,180) to solve the last unknown.

180-70 = 120e^5k

110 = 120e^(5k)

110/120 = e^(5k)

11/12 = e^(5k)

ln(11/12) = 5k

(1/5)ln(11/12) = k

k = -0.0174

T-70 = 120e^(-0.0174t)

Now that the equation has no unknowns besides T, and t, the problem can be answered. Use T=120

120-70 = 120e^(-0.0174t)

50/120 = e^(-0.0174t)

5/12 = e^(-0.0174t)

ln(5/12) = -0.0174t

(1/-0.0174)ln(5/12) = t

t = 50.3143

The coffee is drinkable after 50 minutes.

The next scribe is Sarah

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