A cake removed from the oven has a temperature of 225°f and is left to cool in a

Question

A cake removed from the oven has a temperature of 225°f and is left to cool in a room that has a temperature of 75°f .After 30 min the temperature of the cake is 150°f, what is the temperature of the cake after 60 min?

Over a period of time, a hot object cools to the temperatu TEC (To-C) e , where t is the time it takes for an constant that is associated with the cooling object. A cake 75°F After 30 minutes, the temperature of the cake is 150 After 60 minutes, the temperature of the cake isoF (Do not round until the final answer. Then round to the near oF

Explanation / Answer

C= surrounding temp =75

T0=225

T=75+(225-75)e-kt

After 30 min the temperature of the cake is 150°f,

150=75+(225-75)e-30k

(150-75)/(225-75) =e-30k

0.5=e-30k

-30k=ln(0.5)

k= ln(0.5)/(-30)

k=0.023104906

Thus

T=75+(225-75)e^(-0.023104906t)

T(60) =75+(225-75)e^(-0.023104906*60)

=112.5

Answer will be

112.5 degree Fahrenheit

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