I have a forensics question that asks when a body will reach room temperature if left undisturbed. I am told room temperature is 20 degrees and I am working with Newton's Law of Cooling.
I have an equation:
$$y(t) = 11e^{-0.00265t} + 20$$
This tells me the temperature of the body at time $t$. I want to find the time when the body will reach room temperature.
Just for sake of example, say I want to find the time when the temperature is $25$:
$$\begin{align} 25 &= 11e^{-0.00265t} + 20 \\ (25-20) &= 11e^{-0.00265t} \\ 5 &= 11e^{-0.00265t} \\ \frac{5}{11} &= e^{-0.00265t} \\ \ln{\frac{5}{11}} &= \ln{e^{-0.00265t}} \\ \ln{\frac{5}{11}} &= -0.00265t \\ \frac{\ln{\frac{5}{11}}}{-0.00265} &= t \\ 297.531 &\approx t \end{align}$$
Thus the body will reach $25$ degrees at time $t = 297.531$. This is fine and dandy but when I want to find $t$ when the temperature is $20$, I run into an issue:
$$\begin{align} 20 &= 11e^{-0.00265t}+20 \\ (20-20) &= 11e^{-0.00265t} \\ 0 &= 11e^{-0.00265t} \\ \frac{0}{11} &= e^{-0.00265t} \\ 0 &= e^{-0.00265t} \\ \ln{0} &= \ln{e^{-0.00265t}} \\ \text{Undefined} &= -0.00265t \end{align}$$
Since $\ln{x}$ is only defined in $\mathbb{R}$ for $x \gt 0$.
My equation $y = 11e^{-0.00265x} + 20$ has a horizontal asymptote at $y = 20$. Is this a trick question or there is a solution that can be found using limits? I am familiar with limits with respect to $x$, but not $y$. I need something like $\lim_{y\to20} y = 11e^{-0.00265x}+20$ which doesn't really make sense to me.
The temperature is never $20$. The exponential function is positive for all inputs, as is $11$, so $11\mathrm{e}^{-0.00265t}+20 > 20$ for all $t$. In the limit of infinitely large time, the temperature approaches $20$, getting as close as one likes to $20$ after a sufficiently long time.