Urgent!!! Solve part $BtoG.$

$t=9$mins

Ambient Temperature, $Ta=22C$

Time $tin$ minutes & Liquid temperature $in$ degree celsius $ofT(t)at$ time $t$

$9292$

$10281$

$11270$

$12259$

$13249$

$14239$

$15230$

$16221$

$17213$

$18205$

$19197$

$20189$

$21182$

$22175$

$23169$

$24163$

$25157$

$26151$

$27145$

$28140$

$29135$

$30130$

$31126$

$32121$

$33117$

$34113$

$B.$ You are given the Newton's Law $of$ Cooling $\frac{d}{dt}T(t)=k[T(t)-T_a],to$ model your data,

where $T(t)is$ the temperature $of$ the object $at$ time $t,T_{a}is$ the ambient temperature, and $kis$ the growth constant.

Derive $an$ exponential decay equation $(Hint$: General Solution$)$ from the given Newton's Law $of$ Cooling above using:

$(i)$ Separable Variables method and,

$(20$ marks$)$

$(ii)$ Integration Factor method.

$(20$ marks$)$

$C.$ Using the General Solution from either $B(i)orB(ii)to$ determine the following:

$(i)$ the Particular Solution using the given data.

$(ii)$ the initial temperature $T_i(Hint$: $tat9$ min mark $is$ not the initial time$).$

$D.$ Using the derived exponential decay equation $inC,$ determine how much time $is$ needed for the object $to$ cool $to$ half its initial temperature, $\frac{1}{2}{T}_i.(10$ marks$)$

$E.$ Show mathematically thatk $t_{\text{H}\text{a}\text{l}\text{f}\text{t}\text{i}\text{m}\text{e}}=-ln2,$ where ${?}_{(}$Halftime $),is$ the time taken for the object $to$ cool $to$ half its initial temperature difference.

Hint: Let $T_d(t)=T(t)-T_a,$ where $T_{d}is$ the temperature difference. $(Do$ not use the given numerical data $to$ prove the above$)$

$F.$ Instead $of$ the exponential half$-$time decay expression $k{t}_{\text{H}\text{a}\text{l}\text{t}\text{t}\text{i}\text{m}\text{e}}=-ln2,$ derive the expression for the exponential doubling time growth.

$G.$ State one similarity and one difference between the doubling time and the half time $of$ exponential growt$\frac{h}{d}$ecay equations?