Closed Newton's Law of Cooling Differential Equation

[Ferocity]

Need help guys, pati sa google di ko makita solution neto.
At 9 A.M., a thermometer reading 70°F is taken outdoors where the temperature is
l 5°F. At 9: 05 A.M., the thermometer reading is 45°F. At 9: 10 A.M., the thermometer is
taken back indoors where the temperature is fixed at 70°F. Find (a) the reading at
9: 20 A.M. and (b) when the reading, to the nearest degree, will show the correct
(70°F) indoor temperature.
Thanks in advance sa sasagot.

 

[siMp0lMaN]

credit: google

let 9:10am as t=0t=0 and 70F as my ambient temperature, and α=−ln6115α=−ln⁡6115

now I'll have,

when t=0t=0; x(0)=31.4Fx(0)=31.4F

then my new specific solution is

x(t)=−38.64e(116)−t5+70x(t)=−38.64e(116)−t5+70

the temperature at 9:20am is

since 9:10am is t=0, 9:20am is t=10

x(10)=−38.64e(116)−105+70=58.5Fx(10)=−38.64e(116)−105+70=58.5F

@9:20am x=58.5ºFx=58.5ºF

 

[siMp0lMaN]

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