Respuesta :
Answer:
T ambient = 10 degrees
Explanation:
Using Newton's Law of Cooling:
T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1
Ti = 100
We have two points to evaluate the above equation as follows:
T = 70 @ t = 10 using Eq 1
70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2
T = 50 @ t = 20 using Eq 1
50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3
Solving the above Eq 2 and Eq 3 simultaneously:
Using Eq 2:
(70 - Tamb) / (100 - Tamb) = e^(-10k)
Squaring both sides we get:
((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4
Substitute Eq 4 into Eq 3
50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2
After simplification:
50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)
5000 - 50*Tamb = 4900 - 40*Tamb
Tamb = 100 / 10 = 10 degrees
The ambient temperature will be 10°C. The rate of heat loss is proportional to the difference between the body and ambient temperature.
What does Newton's Law of Cooling state:
The rate of heat loss is proportional to the difference between the body and ambient temperature.
[tex]T_t = T_{amb} + (T_i - T_{amb})\times e^{-kt}[/tex]
There,
[tex]T_t[/tex] - temperature at time t
[tex]T_{amb}[/tex] - ambient temperature
[tex]T_i[/tex] - Initial temperature
[tex]k[/tex] - constant
For the first situation, T = 70°C at t = 10
[tex]70 = T_{amb}+ (100 - T_{amb})\times e^{-10k[/tex]
[tex]\dfrac {70 -T_{amb} } {(100 - T_{amb})}= e^{-10k[/tex]..........2
For second situation, T = 50°C at t = 20.
[tex]50 = T_{amb}+ (100 - T_{amb})\times e^{-20k[/tex].....................3
From solving equation 2, we get,
[tex]T_{amb} = 10^oC[/tex]
Therefore, the ambient temperature will be 10°C.
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