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QUESTION

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

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.