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?

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.

   (  Integrating the expression we will have )

T = Tm +   ---- ( 1 )

where k = time

at T = 0 equation becomes

65 = Tm + C

∴ C = 65 - Tm

Back to equation ( 1 )

T = Tm + ( 65 -Tm )  ---- ( 2 )

After 1/2 minute equation 2 becomes

110 = Tm + ( 65 - Tm ) ---- ( 3 )

After 1 minute equation 2 becomes

140 = Tm + ( 65 - Tm )  ---- ( 4 )

Next step : Divide equation 4 by equation 3

 ----- ( 5 )

Also From equation 3

 --- ( 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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The temperature of the oven is 200°F

Explanation:

Given-

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

On Integrating both sides, we get

On putting the value,

T(0) = 65°F

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

               - 1

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

                    - 2

Dividing equation 2 by 1:

                               - 3

From 1 we have,

Putting this vale in eqn 3. We get,

Therefore, the temperature of the oven is 200°F

your answer is c! hope this : )