On a cold winter night, Anna makes a cup of hot cocoa. When she takes the
first sip, she realizes that the cocoa is too hot to drink. She drops a thermometer into the
cup and discovers that the cocoa is 170° F. After 20 minutes, she tries again and the
cocoa is still too hot at 130°F. She assumes the cocoa will be ready to drink when it is
110°F. If the room temperature is 70° F, how long will it take for the cocoa to cool from
170° F to 110° F? Use Newton's Law of Cooling.
I don't understand how you would get your answer if it doesn't give you the accurate temperature
To determine how long it will take for the cocoa to cool from 170°F to 110°F, we can use Newton's Law of Cooling. This law states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the surrounding temperature.
The formula for Newton's Law of Cooling is:
dT/dt = -k(T - Ts)
Where:
dT/dt is the rate of change of temperature with respect to time.
k is the cooling constant.
T is the temperature of the object.
Ts is the surrounding temperature.
In this case, the surrounding temperature (room temperature) is 70°F, the initial temperature of the cocoa is 170°F, and the final desired temperature is 110°F. We need to find the time it takes for the cocoa to cool down from 170°F to 110°F.
Let's consider T as the temperature of the cocoa at any given time t. The rate of change of temperature with respect to time can be written as:
dT/dt = -k(T - 70)
Since we are interested in the time it takes to cool from 170°F to 110°F (ΔT = 170 - 110 = 60°F), we can rearrange the equation to solve for k:
dT/(T - 70) = -k dt
Integrating both sides of the equation over the temperature range T = 170°F to T = 110°F and t = 0 to t = T will give us:
∫ dT/(T - 70) = -k ∫ dt
ln|T - 70| = -kt + C
where C is the integration constant.
Next, we can apply the initial condition T(0) = 170°F, which means at t = 0, the temperature of the cocoa was 170°F:
ln|170 - 70| = -k(0) + C
ln(100) = C
Substituting ln|T - 70| = -kt + ln(100) back into the equation, we have:
ln|T - 70| = -kt + ln(100)
Now we can find the time it takes for the cocoa to cool from 170°F to 110°F by substituting T = 110°F and T = 170°F into the equation and solving for t:
ln|110 - 70| = -k(t) + ln(100)
ln(40) = -k(t) + ln(100)
ln(40) - ln(100) = -kt
Using logarithmic properties, we can simplify this equation further:
ln(40/100) = -kt
ln(2/5) = -kt
Finally, solving for t:
t = - ln(2/5) / k
The value of k depends on the specific circumstances, such as the material and shape of the cup, conductivity, etc., so it would require more information to determine the exact value of k. However, once the value of k is known, you can substitute it into the equation to find how long it will take for the cocoa to cool from 170°F to 110°F.