Pre Calculus Question

Question

Milk is poured into a glass when it is 33ºF, and it warms to 38ºF in a room that is 75ºF in 10 minutes. What is the temperature after 30 min? 60 min? How long before the milk warms to 60ºF?

Show the process.

Answer 1

Let

T_env = 75°F be the temperature of the room,

T(t) be the temperature of the milk after time t.

We are given

T(0) = 33°F

T(10) = 38°F

We are being asked to calculate T(30), T(60), and the time t₁ at which

T(t₁) = 60°F.

I assume you are to use Newton's Law of Cooling

T(t) = T_env + (T(0) - T_env )e^-rt

for some coefficient of heat transfer r.

We do not know the coefficient r, but we can calculate it from the give T(0) and T(10).

T(10) = T_env +(T(0) - T_env )e^-10r

Solve for r:

ln(T_env - T(10)) = ln(T_env - T(0)) - 10r

r = (ln(T_env - T(0)) - ln(T_env - T(10)))/10 = ln((T_env - T(0)/(T_env - T(10))/10

Substitute actual numbers

r = ln(42/37)/10 = 0.013

Now we can use the Newton's formula to calculate

T(30) = 75 - 42e^-0.013׳⁰ = 46.5°F

T(60) = 75 - 42e^-0.013×⁶⁰ = 55.7°F

T(t₁) = 60

75 - 42e^-rt1 = 60

e^-rt1 = 15/42

-rt₁ = ln(15/42)

t₁ = -ln(15/42)/r = -ln(15/42)/0.013 = 79.2 minutes