A container of hot liquid is placed in a freezer that is kept at a constant temperature of 10°F. The initial temperature of the liquid is 160°F. After 5 minutes, the liquid's temperature is 60°F. How much longer will it take for its temperature to decrease to 20°F? (Round your answer to two decimal places.)

I know you have to use Newton's Law of Cooling, but I keep ending up with the wrong answer

T = Ts + (To-Ts)e^-(kt)

Ts = T surrounding = 10
To = initial T = 160

60 = 10 +(150)e^-(5k)

e^-(5k) = .333333
-5k = -1.1
k = .22
so
20 = 10 + 150 e^-.22t

e^-.22 t = .0666666
.22t = 2.71
t = 12.3 min
12.3 - 5 = 7.3 min more

Thank you so much I realized I was missing the -5 at the end

Ah, Newton's Law of Cooling, a chilling subject indeed! Now, let's cool things down by tackling this temperature dilemma.

First, we need to find the rate at which the liquid is cooling. We can do this by using the formula:

dT/dt = -k(T - T_a)

Where dT/dt represents the rate of change of temperature, k is the cooling constant, T is the temperature of the liquid, and T_a is the temperature of the surrounding environment (in this case, 10°F).

Now we can find the cooling constant, k. We know that after 5 minutes, the temperature dropped from 160°F to 60°F. Putting these values in our formula, we get:

-100 = -k(60 - 10)

Simplifying, we have:

-100 = -50k
k = 2

Great! Now we have our cooling constant. Next, we can use the same formula to find out how much longer it will take for the temperature to decrease to 20°F:

-20 = -2(20 - 10)

Simplifying further, we have:

-20 = -20

Wait a minute, it seems like we have an equation that's already perfectly balanced! That means the liquid will reach 20°F in exactly 0 minutes!

So, the punchline is that the temperature will decrease to 20°F instantly, making it the coolest joke in town! Keep it cool!

To use Newton's Law of Cooling, we need to know the rate at which the temperature of the liquid is decreasing. The law states that the rate of temperature change is proportional to the temperature difference between the object and the surrounding environment.

Let's denote:
- T(time) as the temperature of the liquid at any given time t
- T(freezer) as the temperature of the freezer (10°F in this case)
- T(initial) as the initial temperature of the liquid (160°F)
- k as the cooling constant (which we need to determine)

According to Newton's Law of Cooling, we can express the rate of temperature change as follows:

dT/dt = -k(T - T(freezer))

Where dT/dt is the derivative of T with respect to time t.

Since we know the initial temperature and the temperature after 5 minutes, we can set up an equation using the above formula:

(dT/dt) | t=5 = -k(T(5) - T(freezer))

Plugging in the given values:
- T(5) = 60°F
- T(freezer) = 10°F

We can solve for k using the equation above:

- k = (dT/dt) | t=5 / (T(5) - T(freezer))

Given that the temperature needs to decrease to 20°F, we need to find the time t at which T(t) = 20°F. We can rewrite the formula above in terms of t:

k = (dT/dt) | t=5 / (T(5) - T(freezer))
k = (dT/dt) | t=5 / (60 - 10)

Now, we integrate both sides of the equation to find the equation for temperature T(t):

∫(1/T - 1/T(freezer)) dT = -k∫dt

Integrating the left side:

ln(T) - ln(T(freezer)) = -kt + C

Applying the initial condition T(0) = T(initial):

ln(T(initial)) - ln(T(freezer)) = C

Simplifying the equation:

ln(T) - ln(T(freezer)) = -kt + ln(T(initial)) - ln(T(freezer))
ln(T) = -kt + ln(T(initial))

Now, we plug in the values to solve for k:

ln(20) = -k(t) + ln(160)

Solving for k:
k = (ln(20) - ln(160)) / t

Substituting this value of k into the earlier equation, we get:

k = (ln(20) - ln(160)) / t
(ln(20) - ln(160)) / t = (dT/dt) | t=5 / (60 - 10)

Now, we can find t by rearranging the equation:

t = ((60 - 10) * (ln(20) - ln(160))) / (dT/dt) | t=5

Plugging in the values:
- (dT/dt) | t=5 = (60 - 10) / 5 = 10°F/min

Calculating:

t = ((60 - 10) * (ln(20) - ln(160))) / 10

Evaluating this expression will give you the time it takes for the liquid to decrease to 20°F.