- The best essay writing company you will ever find online
- +1 (510) 327 2058
- support@bestessayswriters.com

Yes if you know the room temperature and the temperature of the cooling object at two known times ##t_{2}>t_{1}>0## after the initial time ##t_{0}=0##.

I will do this in the general case. Somebody else may like to include a specific example (with specific numbers).

Suppose the steady room temperature is ##T_{room}## the temperature of the cooling object at time ##t_{1}>0## is ##H_{1}## where ##T_{room} < H_{1}## and the temperature of the cooling object at time ##t_{2}>t_{1}>0## is ##H_{2}## where ##T_{room} < H_{2} < H_{1}##.
In this situation can be written as giving the temperature ##T## of the object as a function of time ##t## since the object first started cooling in the following way:
##T=f(t)=T_{room}+C*e^{ -k*t}## where the constants ##C>0## and ##k>0## must be found from the given data (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 solve this system of equations you could solve the first equation for ##C## in terms of ##k## to get ##C=(H_{1}-T_{room})*e^{k*t_{1}}## and the substitute 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 solved 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 here would both be negative but the fraction would be positive)

We can then ultimately solve 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 solve for the initial temperature ##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 course if we had specific numbers. But this shows that it is possible to do it in general.

We use cookies to ensure that we give you the best experience on our website. If you continue to use this site we will assume that you are happy with it.Ok