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Energy - Temperature and Temperature Scales

Grade 9IB

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

🔑Concepts

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Temperature is a physical quantity that expresses hot and cold. It is a measure of the average kinetic energy of the particles in a system.

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Internal Energy is the total energy of all particles in an object (kinetic + potential), whereas Temperature only relates to the average kinetic energy.

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The Kelvin scale (KK) is the absolute temperature scale. 0 K0\ K is known as absolute zero, the theoretical temperature where all molecular motion stops.

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The Celsius scale (∘C^{\circ}C) is defined by the freezing point of pure water at 0 ∘C0\ ^{\circ}C and the boiling point at 100 ∘C100\ ^{\circ}C at standard atmospheric pressure.

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A change in temperature of 1 K1\ K is equal in magnitude to a change of 1 ∘C1\ ^{\circ}C. Therefore, ΔTK=ΔTC\Delta T_{K} = \Delta T_{C}.

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Thermal equilibrium is reached when two objects in contact no longer exchange net heat energy because they have reached the same temperature.

📐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:

The surface of the Sun is approximately 5778 K5778\ K. What is this temperature in Celsius?

Solution:

T(∘C)=5778−273.15T(^{\circ}C) = 5778 - 273.15 T(∘C)=5504.85 ∘CT(^{\circ}C) = 5504.85\ ^{\circ}C

Explanation:

To find the Celsius temperature from Kelvin, we use the subtraction constant 273.15273.15 (often rounded to 273273 in introductory courses).

Problem 2:

An object's temperature increases by 45 K45\ K. What is the increase in its temperature in degrees Celsius?

Solution:

ΔT(∘C)=ΔT(K)\Delta T(^{\circ}C) = \Delta T(K) ΔT(∘C)=45 ∘C\Delta T(^{\circ}C) = 45\ ^{\circ}C

Explanation:

Since the size of one unit on the Kelvin scale is identical to the size of one degree on the Celsius scale, any change in temperature (ΔT\Delta T) is the same for both scales.

Problem 3:

Convert 20 ∘C20\ ^{\circ}C to Fahrenheit.

Solution:

T(∘F)=(95×20)+32T(^{\circ}F) = \left( \frac{9}{5} \times 20 \right) + 32 T(∘F)=36+32T(^{\circ}F) = 36 + 32 T(∘F)=68 ∘FT(^{\circ}F) = 68\ ^{\circ}F

Explanation:

Using the conversion formula, we first multiply the Celsius value by 1.81.8 (which is 95\frac{9}{5}) and then add 3232 to account for the different zero points of the scales.