At 2 pm a hot coal was pulled out of a furnace and allowed t

At 2 p.m., a hot coal was pulled out of a furnace and allowed to cool at room temperature (75

Solution

according to newton\'s law of cooling

T(t)=T_a+(T₀-T_a)e^-kt

where T(t) =temperature at time t

T_a=ambiant temperature

T₀=initial temperature

k=constant

435=75+(x-75)e^-k/6

363=75+(x-75)e^-k/3

and on solving these 2 equation we will get

x=525F (required temperature of the furnace)

and

k=-6(2log(2)-log(5))

k=-6(.602-.698)=0.576

(b)

when the coal temperature is 95F

then

95=75+450e^-0.576t

20/450=e^-0.576t

2/45=e^-0.576t

0.0444=e^-0.576t

on solving you will get the time