Question

Exponential law of heating and cooling

Answer 1

A) k = 0.037

B) 186°F

Explanation:

List initial (T₀) and air (Tₐ) temperature

`T_0=280^\circ\\T_a=67^\circ\\`

Part A

`T=T_a+(T_0-T_a)e^(-kt)\\193^\circ=67^\circ+(208^\circ-67^\circ)e^(-k(3))\\193=67+141e^(-3k)\\126=141e^(-3k)\\(126)/(141)=e^(-3k)\\\ln((126)/(141))=-3k\\-(1)/(3)\ln((126)/(141))=k\\k\approx0.037`

Part B

`T=T_a+(T_0-T_a)e^(-kt)\\T=67+(208-67)e^(-0.037(4.5))\\T\approx186^\circ\text{F}`