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Suppose that an object with an initial temperature 𝑇0 is placed in an atmosphere of temperature π‘‡π‘Ž. Newton’s Law of Cooling states that the temperature of the object after 𝑑 minutes is given by the function 𝑇(𝑑) = π‘‡π‘Ž + (𝑇0 βˆ’ π‘‡π‘Ž )𝑒 βˆ’π‘˜π‘‘, where π‘˜ is a positive constant that depends on the type of object. a. A restaurant serves a cup of coffee at a temperature of 180°𝐹. If the manager of the restaurant keeps the thermostat set at 67°𝐹, what will be the temperature of the coffee after 30 minutes if no one drinks it? (Assume that π‘˜ = 0.065.) Round to the nearest whole number and include units in your answer. b. At the same restaurant, a bowl of chili is served at 185°𝐹. If a customer cannot eat it at a temperature higher than 106°𝐹, how long will he need to wait? (Assume that π‘˜ = 0.13.) Round to the nearest whole number and include units in your answer.

 Dec 16, 2020
 #1
avatar+114325 
+1

Ending temp  =  room temp +  ( beginning temp of object  - room temp ) * e-kt

 

Ending temp =  67  +  ( 180  - 67)e (-,065 * 30)

 

Ending temp   = 67   +  ( 113)e(-.065 * 30) 

 

Ending temp  β‰ˆ  83Β° F

 

 

cool cool cool

 Dec 16, 2020
edited by CPhill  Dec 16, 2020
 #2
avatar+114325 
+1

Second one

 

106 =  67  +  ( 185 - 67)e(-..13 * t)     subtract 67 from both sides

 

39  = ( 118)e(-.065 * t)    divide  both sides  by  118

 

(39/118)  = e (-.13 * t)     take the Ln of  both sides

 

Ln ( 39/118)  =Ln  e (-.13 * t)      and we can write

 

Ln ( 39/118)  =  (-.13 * t)  Ln  e        (Ln e  =1....we can ignore this )

 

Ln ( 39/118)  = -.13 * t       divide both sides by -.13

 

Ln (39/118) /-.13  = t  β‰ˆ 9 min

 

 

cool cool cool 

 Dec 16, 2020

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