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Measurement - Measurement of Temperature

Grade 7ICSE

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 a body. It is a scalar quantity.

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The SI unit of temperature is the Kelvin (K\text{K}). Other commonly used units are degree Celsius (∘C^{\circ}\text{C}) and degree Fahrenheit (∘F^{\circ}\text{F}).

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A thermometer works on the principle of thermal expansion of liquids (usually Mercury or Alcohol).

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The Lower Fixed Point (LFP) is the temperature at which pure ice melts at standard atmospheric pressure. For Celsius, it is 0∘C0^{\circ}\text{C}; for Fahrenheit, it is 32∘F32^{\circ}\text{F}.

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The Upper Fixed Point (UFP) is the temperature at which pure water boils at standard atmospheric pressure. For Celsius, it is 100∘C100^{\circ}\text{C}; for Fahrenheit, it is 212∘F212^{\circ}\text{F}.

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Clinical Thermometers are used to measure human body temperature. They have a range from 35∘C35^{\circ}\text{C} to 42∘C42^{\circ}\text{C} and contain a 'kink' (constriction) to prevent the immediate backflow of mercury.

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Laboratory Thermometers are used for scientific experiments and usually have a range from −10∘C-10^{\circ}\text{C} to 110∘C110^{\circ}\text{C}. They do not have a kink.

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Absolute Zero is the theoretical temperature at which molecular motion ceases, equivalent to 0 K0\text{ K} or −273∘C-273^{\circ}\text{C}.

📐Formulae

C100=F−32180\frac{C}{100} = \frac{F - 32}{180}

C5=F−329\frac{C}{5} = \frac{F - 32}{9}

K=C+273K = C + 273

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

💡Examples

Problem 1:

Convert the normal human body temperature, 37∘C37^{\circ}\text{C}, into the Fahrenheit scale.

Solution:

Using the formula: F=(95×C)+32F = (\frac{9}{5} \times C) + 32 F=(95×37)+32F = (\frac{9}{5} \times 37) + 32 F=(9×7.4)+32F = (9 \times 7.4) + 32 F=66.6+32F = 66.6 + 32 F=98.6∘FF = 98.6^{\circ}\text{F}

Explanation:

To convert from Celsius to Fahrenheit, we multiply the Celsius value by 95\frac{9}{5} and then add 3232.

Problem 2:

If the temperature of an object is 313 K313\text{ K}, what is its temperature in Celsius?

Solution:

Using the formula: K=C+273K = C + 273 313=C+273313 = C + 273 C=313−273C = 313 - 273 C=40∘CC = 40^{\circ}\text{C}

Explanation:

To find the Celsius temperature from Kelvin, subtract 273273 from the Kelvin value.

Problem 3:

At what temperature are the Celsius and Fahrenheit scales equal?

Solution:

Let the temperature be xx. Using x5=x−329\frac{x}{5} = \frac{x - 32}{9} 9x=5(x−32)9x = 5(x - 32) 9x=5x−1609x = 5x - 160 4x=−1604x = -160 x=−40x = -40

Explanation:

By setting C=F=xC = F = x in the conversion formula, we find that at −40∘-40^{\circ}, both scales show the same numerical value.