Respuesta :
Answer:
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Step-by-step explanation:
The differential equation governing the given situation is,
[tex]\frac{dT}{dt}=-k\left ( T-T_s \right )[/tex]
Integrating the above we get,
=>[tex]\int \frac{\mathrm{d}T }{T-T_s}=\int -kdt[/tex]
=>[tex]\ln(T-T_s)=-kt+C[/tex]
At t=0 T=T0, applying this condition in the above equation we get
=>[tex]T(t)-T_s=(T(0)-T_s)e^{-kt}[/tex]
It is given that it takes 1 minute for the temperature to fall from 30 to 10.99574 with T(t)=10.99574, T(0)=30 and Ts=10
=>[tex]10.99574-10=(30-10)e^{-kt}[/tex]
=>[tex]0.99574=20e^{-kt}[/tex]
=>[tex]e^{kt}=\frac{20}{0.99574}[/tex]
=>[tex]e^{k\times 1}=\frac{20}{0.99574}[/tex]
=>[tex]k=\ln\frac{20}{0.99574}=3[/tex]
The differential equation that models the temperature in the bar as a function of time is given by:
[tex]\frac{dT_o}{dt} = 4(T_m - T_o)[/tex]
The rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object.
This means that the differential equation is:
[tex]\frac{dT_o}{dt} = k(T_m - T_o)[/tex]
In which:
- [tex]T_m[/tex] is the temperature of the medium.
- [tex]T_o[/tex] is the temperature of the object.
Solving by separation of variables:
[tex]\frac{dT_0}{(T_m - T_o)} = k dt[/tex]
[tex]\int \frac{dT_0}{(T_m - T_o)} = \int k dt[/tex]
[tex]-\ln{(T_m - T_o)} = kt + K[/tex]
[tex]\ln{(T_m - T_o)} = -kt - K[/tex]
[tex]T_m - T_o = -Ke^{-kt}[/tex]
[tex]T_o(t) = T_m + Ke^{-kt}[/tex]
The medium is the room, which has a temperature of 30 degrees Celsius, thus [tex]T_m = 30[/tex]
[tex]T_o(t) = 30 + Ke^{-kt}[/tex]
Initially, the temperature of the metal bar is of 60 degrees, thus [tex]T_o(0) = 60[/tex], and this is used to find K.
[tex]30 + K = 60[/tex]
[tex]K = 30[/tex]
Thus:
[tex]T_o(t) = 30 + Ke^{-kt}[/tex]
[tex]T_o(t) = 30 + 30e^{-kt}[/tex]
[tex]T_o(t) = 30(1 + e^{-kt})[/tex]
One minute later the bar has cooled to 30.54947 degrees, which means that [tex]T_o(1) = 30.54947[/tex], and this is used to find k.
[tex]T_o(t) = 30(1 + e^{-kt})[/tex]
[tex]30.54947 = 30(1 + e^{-k})[/tex]
[tex]1 + e^{-k} = \frac{30.54947}{30}[/tex]
[tex]1 + e^{-k} = 0.0183[/tex]
[tex]e^{-k} = 0.0183[/tex]
[tex]\ln{e^{-k}} = \ln{0.0183}[/tex]
[tex]-k = -4[/tex]
[tex]k = 4[/tex]
Thus, the differential equation is:
[tex]\frac{dT_o}{dt} = 4(T_m - T_o)[/tex]
A similar problem is given at https://brainly.com/question/21046174
