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

tbh i don't know the answer for this question

Answer from: Quest

if i am correct, your answer choices are:

a. the temperature remains constant

b. the temperature increases at a constant rate

c. heat flows at a constant rate into the system

d. heat flows at a constant rate out of the system

e. no heat flows into or out of the system

if so, then the answer is e. no heat flows into or out of the system.

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