Consider a 92∘C cup of coffee placed in a 24∘C room. Suppose it is known that the coffee cools at a rate of 2∘C/min when it is 70∘C. Then according to Newton's law of cooling, the temperature T(t) of the coffee t minutes after being placed in the room satisfies the differential equation,
dT/dt=(2/(70−24))(24−T).
What is the temperature (in Celcius) of the coffee at time t=1 min?
I tried a few methods but I am completely stumped with this questions. Can someone please show me steps on how to do it
To find the temperature of the coffee at time t = 1 minute, we need to solve the given differential equation using separation of variables.
The differential equation is: dT/dt = (2/(70-24))(24-T)
Step 1: Separate the variables:
dT/(24-T) = (2/(70-24))dt
Step 2: Integrate both sides:
∫(1/(24-T)) dT = (2/(70-24)) ∫dt
Step 3: Integrate the left side:
Let u = 24-T, then du = -dT
-∫(1/u) du = (2/(70-24)) ∫dt
- ln|u| = (2/(70-24)) t + C1
Step 4: Simplify the equation:
ln|u| = -(2/(70-24)) t + C1
ln|24-T| = -(2/(70-24)) t + C1
Step 5: Solve for T:
Take the exponential of both sides:
|24-T| = e^(-(2/(70-24)) t + C1)
Since e^C1 is a positive constant, we can use it to eliminate the absolute value:
24-T = ±e^(-(2/(70-24)) t + C1)
Step 6: Determine the positive or negative sign:
Since the coffee is cooling down, we know that the temperature T will be less than 24. Therefore, we choose the negative sign:
24-T = -e^(-(2/(70-24)) t + C1)
Step 7: Determine the value of C1:
At t=0, the initial temperature of the coffee is 92°C. Substitute these values into the equation:
24-92 = -e^C1
-68 = -e^C1
e^C1 = 68
C1 = ln(68)
Step 8: Substitute the value of C1 into the equation:
24-T = -e^(-(2/(70-24)) t + ln(68))
Step 9: Solve for T at t = 1 minute:
Substitute t = 1 into the equation:
24-T = -e^(-(2/(70-24)) + ln(68))
24-T = -e^(-2/46 + ln(68))
24-T = -e^(-1/23 + ln(68))
24-T = -e^(ln(68)/23 - 1/23)
24-T = -e^[(ln(68)-1)/23]
24-T ≈ -0.4462
T ≈ 24 + 0.4462
T ≈ 24.4462°C
Therefore, the temperature of the coffee at time t=1 min is approximately 24.4462°C.
To solve the differential equation and find the temperature of the coffee at t=1 min, we can use Separation of Variables method.
Here are the steps to solve the differential equation:
1. Write the given differential equation:
dT/dt = (2/(70-24))(24-T)
2. Separate the variables by moving the T term to one side and the t term to the other side:
(70-24)/(24-T) dT = 2 dt
3. Integrate both sides with respect to their respective variables:
∫ (70-24)/(24-T) dT = ∫ 2 dt
4. Solve the left side of the equation:
Let u = 24 - T (this is a substitution)
The derivative of u with respect to T is -1:
dT = -du
Substituting into the integral, we get:
∫ (70-24)/u (-du) = - ∫ 2 dt
Simplifying and changing the limits of integration:
-∫ 46/u du = -∫ 2 dt
-46 ln|u| = -2t + C1
5. Solve for u:
ln|u| = (2/46)t + C2
Take the exponential of both sides:
|u| = e^((2/46)t + C2)
Considering the absolute value, we have two possibilities:
u = e^((2/46)t + C2) or u = -e^((2/46)t + C2)
6. Substitute back for u:
Case 1: u = 24 - T
24 - T = e^((2/46)t + C2)
Case 2: u = -(24 - T)
-(24 - T) = e^((2/46)t + C2)
7. Solve for T in each case:
Case 1:
T = 24 - e^((2/46)t + C2)
Case 2:
T = -24 + e^((2/46)t + C2)
8. Use the initial condition given, T(0) = 92°C (initial temperature) when t=0:
T(0) = 24 - e^((2/46)*0 + C2)
92 = 24 - e^(C2)
e^(C2) = -92 + 24
e^(C2) = -68
Therefore, C2 = ln|-68|
9. Substitute C2 back into each case:
Case 1: T = 24 - e^((2/46)t + ln|-68|)
Case 2: T = -24 + e^((2/46)t + ln|-68|)
10. Evaluate the temperature at t=1 min:
T(1) = 24 - e^((2/46)*1 + ln|-68|)
= 24 - e^(1/23 - ln|68|)
= 24 - e^(1/23) / 68
Hence, the temperature of the coffee at t = 1 min is 23.3508 °C (rounded to four decimal places).
They gave you the hint "Newton's Law of Cooling" so you do not even have to do the calculus.
for example:
http://www.softschools.com/formulas/physics/newtons_law_of_cooling_formula/93/