Respuesta :
Answer:
a) True. The differential equation is [tex]\frac{dT}{dt}+0.0004\cdot T = 0.06[/tex], where [tex]T[/tex] is measured in Celsius.
b) The temperature of the object after 200 second inside the microwave is approximately 34.611 ºC.
Step-by-step explanation:
a) The object is heated up by convection and radiation, from First Law of Thermodynamics we get that energy interactions on the object are represented by:
[tex]-\dot Q_{out} + \dot W_{in} = \frac{dU_{sys}}{dt}[/tex] (Eq. 1)
Where:
[tex]\dot Q_{in}[/tex] - Heat transfer rate to the object, measured in watts.
[tex]\dot W_{in}[/tex] - Electric power given by the microwave owen, measured in watts.
[tex]U_{sys}[/tex] - Internal energy of the object, measured in watts.
Then, we expand the expression by means of definition from Heat Transfer and Thermodynamics:
[tex]-h_{c}\cdot A_{s}\cdot (T-T_{\infty})+\dot W_{in} = m\cdot c\cdot \frac{dT}{dt}[/tex]
Where:
[tex]h_{c}[/tex] - Heat transfer coefficient, measured in watts per square meter-Celsius.
[tex]A_{s}[/tex] - Surface area of the object, measured in square meters.
[tex]T[/tex] - Temperature of the object, measured in Celsius.
[tex]T_{\infty}[/tex] - Internal temperature of the microwave oven, measured in Celsius.
[tex]m[/tex] - Mass of the object, measured in kilograms.
[tex]c[/tex] - Specific heat of the object, measured in joules per kilogram-Celsius.
After some algebraic handling, we get this non-homogeneous first order differential equation:
[tex]\frac{dT}{dt} = -\frac{h_{c}\cdot A_{s}}{m\cdot c} \cdot (T-T_{\infty}) + \frac{\dot W_{in}}{m\cdot c}[/tex]
[tex]\frac{dT}{dt}+\frac{h_{c}\cdot A_{s}}{m\cdot c}\cdot T =\frac{h_{c}\cdot A_{s}}{m\cdot c}\cdot T_{\infty}+\frac{\dot W_{in}}{m\cdot c}[/tex] (Ec. 2)
If we know that [tex]h_{c} = 40\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex], [tex]A_{s} = 0.01\,m^{2}[/tex], [tex]m = 1\,kg[/tex], [tex]c = 1000\,\frac{J}{kg\cdot ^{\circ}C}[/tex], [tex]T_{\infty} = 25\,^{\circ}C[/tex] and [tex]\dot W_{in} = 50\,W[/tex], the differential equation is:
[tex]\frac{dT}{dt}+0.0004\cdot T = 0.06[/tex]
b) The solution of this differential equation is:
[tex]T(t) = T_{o}\cdot e^{-0.0004\cdot t}+ \frac{0.06}{0.0004}\cdot (1-e^{-0.0004\cdot t})[/tex]
Where [tex]T_{o}[/tex] is the initial temperature of the object, measured in Celsius.
[tex]T(t) = T_{o}\cdot e^{-0.0004\cdot t}+150\cdot (1-e^{-0.0004\cdot t})[/tex]
[tex]T(t) = 150+(T_{o}-150)\cdot e^{-0.0004\cdot t}[/tex]
Where [tex]t[/tex] is the time, measured in seconds.
If we know that [tex]T_{o} = 25\,^{\circ}C[/tex], then:
[tex]T(t) = 150-125\cdot e^{-0.0004\cdot t}[/tex] (Ec. 3)
For [tex]t = 200\,s[/tex], we get that temperature of the object is:
[tex]T(200) = 150-125\cdot e^{-0.0004\times 200}[/tex]
[tex]T(200) \approx 34.611\,^{\circ}C[/tex]
The temperature of the object after 200 second inside the microwave is approximately 34.611 ºC.
