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)
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)
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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.
Updated On: Jul 27, 2026
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Verified By Collegedunia
Correct Answer: 334.9
Solution and Explanation
Step 1: Recall the working of a Kata thermometer.
A Kata thermometer is an alcohol-in-glass thermometer used to measure the cooling power of mine air, which combines the effects of air temperature, humidity and velocity into one number. It is first warmed above a marked upper temperature and then allowed to cool in the air being tested. The time it takes for the alcohol column to fall between two marked temperatures is recorded.
Each thermometer carries its own Kata factor F, stamped on the stem, which represents the heat lost per unit area of the bulb while the alcohol falls through that marked range.
Step 2: Write the formula for dry Kata cooling power.
The dry cooling power H is the heat lost per unit area per second, found by dividing the Kata factor by the time taken for the fall.
\[ H = \frac{F}{t} \]
where F is in milli-calories per cm\(^2\) and t is in seconds, so H comes out in milli-calories cm\(^{-2}\) s\(^{-1}\).
Step 3: Substitute the given values.
Here \(F = 480\) milli-calories cm\(^{-2}\) and \(t = 60\) s (the time for the alcohol to fall from \(38^{\circ}\)C to \(35^{\circ}\)C).
\[ H = \frac{480}{60} = 8 \text{ milli-cal cm}^{-2}\text{s}^{-1} \]
Step 4: Convert milli-calories per cm squared per second into watts per m squared.
One calorie equals 4.186 J, so one milli-calorie equals \(4.186 \times 10^{-3}\) J.
\[ H = 8 \times 4.186 \times 10^{-3} \text{ J cm}^{-2}\text{s}^{-1} = 0.033488 \text{ W cm}^{-2} \]
One m\(^2\) has \(10^4\) cm\(^2\), so multiply by \(10^4\) to convert cm\(^2\) to m\(^2\).
\[ H = 0.033488 \times 10^4 = 334.88 \text{ W m}^{-2} \]
Final Answer:
\[ \boxed{H = 334.9 \text{ W m}^{-2}} \]
A Kata thermometer is an alcohol-in-glass thermometer used to measure the cooling power of mine air, which combines the effects of air temperature, humidity and velocity into one number. It is first warmed above a marked upper temperature and then allowed to cool in the air being tested. The time it takes for the alcohol column to fall between two marked temperatures is recorded.
Each thermometer carries its own Kata factor F, stamped on the stem, which represents the heat lost per unit area of the bulb while the alcohol falls through that marked range.
Step 2: Write the formula for dry Kata cooling power.
The dry cooling power H is the heat lost per unit area per second, found by dividing the Kata factor by the time taken for the fall.
\[ H = \frac{F}{t} \]
where F is in milli-calories per cm\(^2\) and t is in seconds, so H comes out in milli-calories cm\(^{-2}\) s\(^{-1}\).
Step 3: Substitute the given values.
Here \(F = 480\) milli-calories cm\(^{-2}\) and \(t = 60\) s (the time for the alcohol to fall from \(38^{\circ}\)C to \(35^{\circ}\)C).
\[ H = \frac{480}{60} = 8 \text{ milli-cal cm}^{-2}\text{s}^{-1} \]
Step 4: Convert milli-calories per cm squared per second into watts per m squared.
One calorie equals 4.186 J, so one milli-calorie equals \(4.186 \times 10^{-3}\) J.
\[ H = 8 \times 4.186 \times 10^{-3} \text{ J cm}^{-2}\text{s}^{-1} = 0.033488 \text{ W cm}^{-2} \]
One m\(^2\) has \(10^4\) cm\(^2\), so multiply by \(10^4\) to convert cm\(^2\) to m\(^2\).
\[ H = 0.033488 \times 10^4 = 334.88 \text{ W m}^{-2} \]
Final Answer:
\[ \boxed{H = 334.9 \text{ W m}^{-2}} \]
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