Answer:
Step-by-step explanation:
The rate of change of temperature of the coffee with respect to time is expressed by the differential equation
dT/dt=k(T−A) when k = -1 and A = 23. The equation will become:
dT/dt = -(T-23)
If the coffee cools at the rate of 1°C per minute, then dT/dt = -1 and T = 70
Substituting into the equation:
-1 = k(70-23)
-1 = 47k
k = -1/47
k = -0.0213
Substituting k = -0.0213 into the original equation, the differential equation will be:
dT/dt=k(T−A)
dT/dt = -0.0213(T-23)
dT/dt = -0.0213T+0.489
To get the value of T, we will use variable separable method
dt = dT/-0.0213T+0.489
Integrate both sides
t = -0.0213ln(-0.0213T+0.489)
At t = 1 minute
1 = -0.0213ln(-0.0213T+0.489)
1/-0.0213 = ln(-0.0213T+0.489)
-46.95 = ln(-0.0213T+0.489)
Apply exp to both sides
e^-46.95 = e^ln(-0.0213T+0.489)
4.073×10^-21= -0.0213T+0.489
-0.0213T = 4.073×10^-21-0.489
-0.0213T = -0.489
T = 0.489/0.0213
T = 22.96°C
