Physics, 04.03.2020 19:25 sannai0415

A thermometer reading 65° F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 110° F after 1 2 minute and 140° F after 1 minute. How hot is the oven?

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Answer from: Raquelerol

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.

$\frac{dT}{dt} = K(T -Tm )$ ( Integrating the expression we will have )

T = Tm + $Ce^{kt}$ ---- ( 1 )

where k = time

at T = 0 equation becomes

65 = Tm + C

∴ C = 65 - Tm

Back to equation ( 1 )

T = Tm + ( 65 -Tm ) $e^{kt}$ ---- ( 2 )

After 1/2 minute equation 2 becomes

110 = Tm + ( 65 - Tm ) $e^{0.5k}$ ---- ( 3 )

After 1 minute equation 2 becomes

140 = Tm + ( 65 - Tm ) $e^{k}$ ---- ( 4 )

Next step : Divide equation 4 by equation 3

$e^{0.5k} = \frac{140 - Tm}{110 - Tm}$ ----- ( 5 )

Also From equation 3

$e^{0.5k} = \frac{110 - Tm}{65 - Tm}$ --- ( 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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Answer from: rol15

The temperature of the oven is 200°F

Explanation:

Given-

We have to apply Newton's law of cooling or heating

$\frac{dT}{dt} = k ( T - Tm)\\\\\frac{dT}{T - Tm} = kdt$

On Integrating both sides, we get

T = Tm + Ce^k^t

On putting the value,

T(0) = 65°F

$65 = Tm + C\\C = 65 - Tm\\\\T = Tm + (65 - Tm) e^k^t$

After 1/2 minute, thermometer reads 110°F. So,

110 = Tm + (65 - Tm)e^0^.^5^k - 1

After 1 minute, thermometer reads 140°F. So,

140 = Tm + (65 - Tm)e^k - 2

Dividing equation 2 by 1:

$e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}$ - 3

From 1 we have,

$e^0^.^5^k = \frac{110 - Tm}{65 - Tm}$

Putting this vale in eqn 3. We get,

$\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}$

$(110-Tm)^2 = (140-Tm) (65-Tm)\\\\12100 + Tm^2 - 220Tm = 9100 - 140Tm - 65Tm + Tm^2\\\\3000 - 220Tm = -205Tm\\\\3000 = 15Tm\\\\Tm = 200$

Therefore, the temperature of the oven is 200°F

Answer from: Quest

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.

Answer from: Quest

What even is this. where is the question

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