Respuesta :
The temperature of the oven is 200°F
Explanation:
Given-
We have to apply Newton's law of cooling or heating
[tex]\frac{dT}{dt} = k ( T - Tm)\\\\\frac{dT}{T - Tm} = kdt[/tex]
On Integrating both sides, we get
[tex]T = Tm + Ce^k^t[/tex]
On putting the value,
T(0) = 65°F
[tex]65 = Tm + C\\C = 65 - Tm\\\\T = Tm + (65 - Tm) e^k^t[/tex]
After 1/2 minute, thermometer reads 110°F. So,
[tex]110 = Tm + (65 - Tm)e^0^.^5^k[/tex] - 1
After 1 minute, thermometer reads 140°F. So,
[tex]140 = Tm + (65 - Tm)e^k[/tex] - 2
Dividing equation 2 by 1:
[tex]e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}[/tex] - 3
From 1 we have,
[tex]e^0^.^5^k = \frac{110 - Tm}{65 - Tm}[/tex]
Putting this vale in eqn 3. We get,
[tex]\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}[/tex]
[tex](110-Tm)^2 = (140-Tm) (65-Tm)\\\\12100 + Tm^2 - 220Tm = 9100 - 140Tm - 65Tm + Tm^2\\\\3000 - 220Tm = -205Tm\\\\3000 = 15Tm\\\\Tm = 200[/tex]
Therefore, the temperature of the oven is 200°F
The Temperature ( hotness ) of the oven is ; 200°F
Given data :
Initial thermometer reading ( To ) = 65°F
Thermometer reading after 1/2 minute ( T1/2 ) = 110°F
Thermometer reading after 1 minute ( T1 ) = 140°F
Temperature of oven ( Tm ) = ?
Determine the Temperature of the Oven
To determine the temperature of the oven we will apply Newton's law of heating and cooling.
[tex]\frac{dT}{dt} = K(T -Tm )[/tex] ( Integrating the expression we will have )
T = Tm + [tex]Ce^{kt}[/tex] ---- ( 1 )
where k = time
at T = 0 equation becomes
65 = Tm + C
∴ C = 65 - Tm
Back to equation ( 1 )
T = Tm + ( 65 -Tm )[tex]e^{kt}[/tex] ---- ( 2 )
After 1/2 minute equation 2 becomes
110 = Tm + ( 65 - Tm )[tex]e^{0.5k}[/tex] ---- ( 3 )
After 1 minute equation 2 becomes
140 = Tm + ( 65 - Tm )[tex]e^{k}[/tex] ---- ( 4 )
Next step : Divide equation 4 by equation 3
[tex]e^{0.5k} = \frac{140 - Tm}{110 - Tm}[/tex] ----- ( 5 )
Also From equation 3
[tex]e^{0.5k} = \frac{110 - Tm}{65 - Tm}[/tex] --- ( 6 )
Next step : Equating equations ( 5 ) and ( 6 )
( 110 - Tm )² = ( 140 - Tm )( 65 - Tm )
3000 - 220 Tm = -205Tm
∴ Tm = 3000 / 15 = 200°F
Hence we can conclude that the temperature ( hotness ) of the oven is 200°F.
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