Can Newton’s Law of Cooling be used to find an initial temperature?

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Yes, if you apprehend the capacity sky and the sky of the composure motive at two apprehendn opportunitys ##t_{2}>t_{1}>0## succeeding the judicious opportunity ##t_{0}=0##.

I conquer do this in the public plight. Somebody else may love to conceive a inequitable issue (delay inequitable aggregate).

Suppose the well-regulated capacity sky is ##T_{room}##, the sky of the composure motive at opportunity ##t_{1}>0## is ##H_{1}##, wless ##T_{room} < H_{1}##, and the sky of the composure motive at opportunity ##t_{2}>t_{1}>0## is ##H_{2}##, wless ##T_{room} < H_{2} < H_{1}##.

In this top, can be written as giving the sky ##T## of the motive as a employment of opportunity ##t## since the motive primitive launched composure in the aftercited way:

##T=f(t)=T_{room}+C*e^{ -k*t}##, wless the constants ##C>0## and ##k>0## must be base from the loving basis (that ##f(t_{1})=H_{1}## and ##f(t_{2})=H_{2}##).

This gives two equations in two unknowns (##C## and ##k## are the unknowns): ##H_{1}=T_{room}+C*e^{ -k*t_{1})## and ##H_{2}=T_{room}+C*e^{ -k*t_{2}}##.

To rerework-out this arrangement of equations, you could rerework-out the primitive equation for ##C## in conditions of ##k## to get ##C=(H_{1}-T_{room})*e^{k*t_{1}}## and the represent this into the second equation to get

##H_{2}=T_{room}+(H_{1}-T_{room}) * e^{k*(t_{1}-t_{2})}##

This equation can then be resolved for ##k## by using logarithms:

##e^{k*(t_{1}-t_{2})}=(H_{2}-T_{room})/(H_{1}-T_{room})##

##Rightarrow k=(ln((H_{2}-T_{room})/(H_{1}-T_{room})))/(t_{1}-t_{2})=(ln(H_{2}-T_{room})-ln(H_{1}-T_{room}))/(t_{1}-t_{2})## (the numerator and denominator less would twain be denying, but the concern would be indisputable)

We can then however rerework-out for ##C## by substituting into ##C=(H_{1}-T_{room})*e^{k*t_{1}}## to get:

##C=(H_{1}-T_{room})*e^{((ln(H_{2}-T_{room})-ln(H_{1}-T_{room}))/(t_{1}-t_{2}))*t_{1}}##

Finally, you can rerework-out for the judicious sky ##H_{0}## as

##H_{0}=f(0)=T_{room}+Ce^{0}=T_{room}+C##

##=T_{room}+(H_{1}-T_{room})*e^{((ln(H_{2}-T_{room})-ln(H_{1}-T_{room}))/(t_{1}-t_{2}))*t_{1}}##

All this would be easier, of continuity, if we had inequitable aggregate. But this shows that it is practicable to do it in public.


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