Respuesta :
Let T represent the temperature of the object
Let To represent the temperature of the surroundings
Newton’s law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. This can be expressed as
[tex]\frac{dT}{dt}\text{ = }-\text{ k}(T-T_o)[/tex]k is a constant and it is negative because the temperature is reducing. The next step is to separate the variables. We have
[tex]\begin{gathered} \frac{dT}{(T\text{ - To)}}\text{ = - kdt} \\ By\text{ integrating both sides, we have} \\ \int \frac{dT}{T\text{ - To}}\text{ = }\int -\text{ kdt} \\ \ln \text{ }\lvert T-To\rvert\text{ = }-\text{ kt + C} \end{gathered}[/tex]Taking the exponents of both sides, it becomes
[tex]e^{\ln \lvert T-To\rvert}=e^{(-\text{ kt + }C)}[/tex]By applying the rules of logarithms and exponents,
e^ln = 1
a^(m + n) = a^ma^n
Thus, the equation becomes
[tex]\begin{gathered} T-To=e^{-kt}e^C \\ \text{Let e}^C\text{ = A} \\ \text{Thus, we have} \\ T-To=Ae^{-\text{ kt}} \\ T=Ae^{-\text{ kt}}\text{ + To} \end{gathered}[/tex]We would apply this formula in solving the given problem. From the information given,
To = 70, Intial temrperature when time, t = 0 is 200
After 1 min, t = 1, T = 190
We would calculate the value of A and k. We have
For t = 0,
200 = Ae^- (k * 0) + 70
200 = Ae^0 + 70
200 = A + 70
A = 200 - 70 = 130
The equation becomes
T = 130e^- kt + 70
Substituting T = 190 and t = 1, it becomes
190 = 130e^- k(1) + 70
190 = 130e^- k + 70
190 - 70 = 130e^- k
120 = 130e^- k
120/130 = e^- k
Taking the natural log of both sides,
ln (120/130) = ln e^- k
ln e = 1
ln (120/130) = - k
k = ln (120/130)/- 1
k = 0.08
Therefore, the equation would be
[tex]T=130e^{-\text{ 0.08t}}\text{ + 70}[/tex]We want to find t when T = 150. By substituting T = 150 into the formula, we have
150 = 130e^- 0.08t + 70
150 - 70 = 130e^- 0.08t
80 = 130e^- 0.08t
80/130 = e^- 0.08t
Taking the natural log of both sides,
ln 80/130 = ln e^- 0.08t
ln 80/130 = - 0.08t
t = (ln 80/130 )/(- 0.08)
t = 6.07 mins
