Respuesta :
Answer:
70.00143
Step-by-step explanation:
Given that a thermometer reading 100◦F is placed in a medium having a constant temperature of 70◦F. After 6 min, the thermometer reads 80◦F.
We know that Newton law of cooling has formula as
[tex]T(t) = Ts +(T_0 - T_s)e^{-kt.}[/tex]
where Ts is room temperature and To is initial temperature.
Using the fact that when t =6, T (6) = 80
T0 = 100 and Ts =70 we can find k
[tex]80 = 70+(100-70)e^{-6k} \\e^{-6k}=0.33333\\k = 0.1831[/tex]
Using this we find temperature when t=20
[tex]T(20) = 70+(100-70)e^{-20*0.1831}\\T(20) = 70+(100-70)e^{-3.6624}\\\\T(20) = 70.00143[/tex]
The reading of the thermometer after 20 minutes is; T(20) = 70.77°F
We will solve this by making use of Newton's law of cooling Formula which is;
T(t) = T_s + (T_o - T_s)e^(-kt)
Where;
T(t) is temperature at time of t minutes
T_s is constant temperature of surroundings
T_o is initial temperature
k is cooling constant
We are given;
T_s = 70°F
T_o = 100°F
T(6) = 80°F
Thus;
80 = 70 + (100 - 70)e^(-6k)
80 - 70 = 30e^(-6k)
10/30 = e^(-6k)
In 0.3333 = - 6k
-1.0987/-6 = k
k = 0.1831
Thus,at t = 20 minutes;
T(20) = 70 + (100 - 70)e^(-20 × 0.1831)
T(20) = 70 + 30e^(-3.662)
T(20) = 70.77°F
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