Question

In an underground mine atmosphere, the alcohol of a Kata thermometer took \(60\) s to fall from \(38^{\circ}\)C to \(35^{\circ}\)C. The Kata factor of the thermometer is \(480\) milli-calories cm\(^{-2}\). The Kata cooling power, in \(W\,m^{-2}\), is . (Rounded off to one decimal place)

Hint:

Divide the Kata factor by the fall time to get the cooling power in milli-calories per cm squared per second, then convert the units to W per m squared.

Answer 1

Correct Answer: 334.9

We can reach the same cooling power by converting units first and combining them at the end, instead of computing the milli-calorie rate first and converting last. This checks the earlier answer through a different order of working.

The Kata factor tells us the heat lost from one square centimetre of the thermometer's bulb while the alcohol column falls through its marked range. Here that heat is $F = 480$ milli-calories per cm$^2$.

Convert this heat into joules right away, using $1$ calorie $= 4.186$ J, so $1$ milli-calorie $= 4.186\times10^{-3}$ J.

\[ Q_{heat} = 480 \times 4.186\times10^{-3} = 2.00928 \text{ J per cm}^2 \]

Convert the area from cm$^2$ to m$^2$ next. Since $1$ m$^2 = 10^4$ cm$^2$, the heat lost per m$^2$ is:

\[ Q_{heat} = 2.00928 \times 10^4 = 20092.8 \text{ J per m}^2 \]

This is the heat lost per square metre over the whole recorded time of $t = 60$ s (the time for the alcohol to fall from $38^{\circ}$C to $35^{\circ}$C). The cooling power is this heat divided by the time, since power is energy per second.

\[ H = \frac{20092.8}{60} = 334.88 \text{ W m}^{-2} \]

Let's summarize:

• The Kata factor gives heat lost per unit bulb area over the fall time, here in milli-calories.
• Converting to joules and then to a per square metre basis first, then dividing by time, gives the same cooling power as converting at the end.
• The dry Kata cooling power works out to 334.9 W/m$^2$, confirming the direct formula result.

So the Kata cooling power of the air in the underground mine is 334.9 W/m$^2$.