My question refers to the cooling of water in a well-insulated , cylindrical cup that is opened at the top.
$0.336$Kg of water at $84$C$^\circ$ is poured into the cup which is left to stand on a table at an ambient temperature of $25$C$^\circ$. The specific heat capacity of water is $4150$ J Kg$^{-1}$ K$^{-1}$ and the U-value is $10.64$Wm$^{-2}$K$^{-1}$. The area of the water surface that is exposed to the surroundings air is $0.00447$ m$^2$.
So, how to calculate the time taken in seconds for the water to cool down to $46$C$^\circ$?
This is what I have done so far….
$$\begin{align*}
\text{Let } x & = \text{(U-value)(Surface Area)/(mass)(Heat Capacity)}\\
\text{Let } y_1 &= 46\text{C}^\circ (\text{cool down temp})\\
\text{Let } y_2 &= 84\text{C}^\circ (\text{Water temp})\\
\text{Let } y_3 &= 25\text{C}^\circ (\text{Ambient tempt})
\end{align*}$$
Taking time to be $T$, so…
$$\begin{align*}
T & = \dfrac{-1}{x} \log \frac{y_1-y_3}{y_2-y_3}\\
& = -13153 (\text{sec})
\end{align*}$$
I got negative, so my guess something is wrong here. The formula is from my book. Need help….
Best Answer
The formula that you used does not produce a negative number. Note that $\frac{y_2-y_3}{y_1-y_3}$ is between $0$ and $1$, so its logarithm is negative. The minus sign in the front of your formula then produces a positive number.
Remark: Do check what kind of log is involved in the formula. You used logarithm to the base $10$. It may be that logarithm to the base $e$ (natural logarithm) is intended. I cannot tell from the numbers, since the difference can be absorbed into the definition of $U$-value. (Whether you use base $10$ or base $e$, the log of $\frac{21}{59}$ is negative.)