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

Your answer is c. explanation: plasma is the fourth state of matter, derived from the gaseous state of matter. plasma can contain free electrons and/or ions. while plasma is never neutral, it can either be positively or negatively charged. gases, however, are electrically neutral.

Answer from: Quest

a. carbon monoxide from car exhaust

b.particulate matter, such as soot

c.water vapor from a factory smokestack

d.acid rain

these are the answer choices im on that question right now

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A thermometer reading 65° F is placed in an oven preheated to a constant temperature. Through a glas...