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Physics: Thermal Physics - Temperature Scales (Celsius, Kelvin, and Fahrenheit)

Grade 8IB

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

🔑Concepts

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Temperature is a measure of the average kinetic energy of the particles in a substance. It determines the direction of thermal energy flow.

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The Celsius scale (∘C^\circ C) is based on the properties of water. The freezing point of water is defined as 0∘C0^\circ C and the boiling point as 100∘C100^\circ C at standard atmospheric pressure.

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The Kelvin scale (KK) is the SI unit for temperature. It is an absolute scale, meaning it starts at Absolute Zero (0K0 K), the theoretical temperature where all molecular motion stops. There are no negative temperatures on the Kelvin scale.

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The Fahrenheit scale (∘F^\circ F) defines the freezing point of water at 32∘F32^\circ F and the boiling point at 212∘F212^\circ F. This scale is primarily used in the United States.

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A temperature interval of 1K1 K is equivalent to an interval of 1∘C1^\circ C. However, 1∘C1^\circ C is equivalent to an interval of 1.8∘F1.8^\circ F.

📐Formulae

T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15

T(∘C)=T(K)−273.15T(^\circ C) = T(K) - 273.15

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

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

💡Examples

Problem 1:

Convert a comfortable room temperature of 25∘C25^\circ C to the Kelvin scale.

Solution:

T(K)=25+273=298KT(K) = 25 + 273 = 298 K

Explanation:

To convert from Celsius to Kelvin, we add 273273 (the approximate difference between the two scales) to the Celsius reading.

Problem 2:

A liquid is heated to 122∘F122^\circ F. What is this temperature in Celsius (∘C^\circ C)?

Solution:

T(∘C)=59×(122−32)T(^\circ C) = \frac{5}{9} \times (122 - 32) T(∘C)=59×90=50∘CT(^\circ C) = \frac{5}{9} \times 90 = 50^\circ C

Explanation:

Subtract 3232 from the Fahrenheit value to find the offset, then multiply by 59\frac{5}{9} to adjust the scale size.

Problem 3:

If a gas is at 350K350 K, find its temperature in Celsius using vertical subtraction.

Solution:

We use the relation T(∘C)=T(K)−273T(^\circ C) = T(K) - 273: 350−27377\begin{array}{r} 350 \\ -273 \\ \hline 77 \end{array} Thus, the temperature is 77∘C77^\circ C.

Explanation:

To find the Celsius value, we subtract the Absolute Zero constant (273273) from the Kelvin temperature.

Problem 4:

Convert 10∘C10^\circ C to Fahrenheit (∘F^\circ F).

Solution:

T(∘F)=(95×10)+32T(^\circ F) = \left( \frac{9}{5} \times 10 \right) + 32 T(∘F)=18+32=50∘FT(^\circ F) = 18 + 32 = 50^\circ F

Explanation:

Multiply the Celsius temperature by 95\frac{9}{5} (or 1.81.8) and then add 3232 to account for the Fahrenheit freezing point offset.