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Temperature and its Measurement - Air Temperature

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Temperature is defined as the degree of hotness or coldness of an object or the atmosphere. It is measured using an instrument called a thermometer.

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Air temperature refers to the temperature of the air around us. It is a critical component of weather and climate measurement.

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The SI unit of temperature is Kelvin (KK), but in daily life, temperatures are commonly measured in Degree Celsius (∘C^{\circ}C) and Degree Fahrenheit (∘F^{\circ}F).

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On the Celsius scale, the freezing point of water is 0∘C0^{\circ}C and the boiling point is 100∘C100^{\circ}C. On the Fahrenheit scale, these are 32∘F32^{\circ}F and 212∘F212^{\circ}F respectively.

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A Maximum-Minimum Thermometer (also known as Six's thermometer) is specifically used to record the highest and lowest temperatures of the day for meteorological purposes.

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Clinical thermometers used for body temperature usually range from 35∘C35^{\circ}C to 42∘C42^{\circ}C, whereas laboratory thermometers have a wider range, typically from −10∘C-10^{\circ}C to 110∘C110^{\circ}C.

📐Formulae

F=(95×C)+32F = \left( \frac{9}{5} \times C \right) + 32

C=59×(F−32)C = \frac{5}{9} \times (F - 32)

Temperature Range=Maximum Temperature−Minimum Temperature\text{Temperature Range} = \text{Maximum Temperature} - \text{Minimum Temperature}

💡Examples

Problem 1:

The air temperature in a city was recorded as 35∘C35^{\circ}C. Convert this temperature into degrees Fahrenheit (∘F^{\circ}F).

Solution:

Using the formula F=(95×C)+32F = \left( \frac{9}{5} \times C \right) + 32, substitute C=35C = 35: F=(95×35)+32F = \left( \frac{9}{5} \times 35 \right) + 32 F=(9×7)+32F = (9 \times 7) + 32 F=63+32F = 63 + 32 F=95∘FF = 95^{\circ}F

Explanation:

To convert Celsius to Fahrenheit, multiply the Celsius value by 95\frac{9}{5} (or 1.81.8) and then add 3232 to the result.

Problem 2:

On a particular summer day, the maximum air temperature was 42∘C42^{\circ}C and the minimum air temperature was 29∘C29^{\circ}C. Calculate the temperature range for that day.

Solution:

To find the range, subtract the minimum temperature from the maximum temperature: 42−2913\begin{array}{r} 42 \\ - 29 \\ \hline 13 \end{array}

The temperature range is 13∘C13^{\circ}C.

Explanation:

The range of temperature is the difference between the highest (maximum) and lowest (minimum) temperatures recorded over a specific period, usually 2424 hours.

Problem 3:

If a weather station reports a temperature of 113∘F113^{\circ}F, what is the equivalent temperature in degrees Celsius (∘C^{\circ}C)?

Solution:

Using the formula C=59×(F−32)C = \frac{5}{9} \times (F - 32), substitute F=113F = 113: C=59×(113−32)C = \frac{5}{9} \times (113 - 32) C=59×81C = \frac{5}{9} \times 81 C=5×9C = 5 \times 9 C=45∘CC = 45^{\circ}C

Explanation:

To convert Fahrenheit to Celsius, first subtract 3232 from the Fahrenheit value, then multiply the result by 59\frac{5}{9}.