# Time to take for water to cool down.

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....

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## 1 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.)

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I saw my mistake. Its log base e....Thanks for the help =D – Jack Apr 14 '12 at 9:11
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