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Physics - Heat and Energy (Temperature, Anomalous Expansion, Energy Sources)

Grade 9ICSE

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

🔑Concepts

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Heat is the total internal energy of the molecules constituting a body, measured in Joules (JJ) or calories (calcal), where 1 cal≈4.186 J1\text{ cal} \approx 4.186\text{ J}.

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Temperature is the degree of hotness or coldness of a body, determining the direction of flow of heat. Its SI unit is Kelvin (KK).

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Linear expansion coefficient (α\alpha) is the increase in length per unit length per degree rise in temperature: α=ΔLL0ΔT\alpha = \frac{\Delta L}{L_0 \Delta T}.

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Anomalous expansion of water: Water contracts when heated from 0∘C0^{\circ}C to 4∘C4^{\circ}C. It has the minimum volume and maximum density (1000 kg m−31000\text{ kg m}^{-3}) at 4∘C4^{\circ}C.

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The Hope's Experiment demonstrates the anomalous expansion of water and explains how aquatic life survives in frozen ponds as the bottom layer remains at 4∘C4^{\circ}C.

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Greenhouse Effect: The process by which infrared radiations of long wavelengths are trapped by atmospheric gases like CO2CO_2, CH4CH_4, and N2ON_2O, leading to Global Warming.

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Renewable energy sources are inexhaustible (e.g., Solar, Wind, Hydro), whereas non-renewable sources are exhaustible (e.g., Coal, Petroleum, Natural Gas).

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Nuclear Energy: Energy released during nuclear fission (splitting of heavy nuclei like 235U^{235}U) or nuclear fusion (combining of light nuclei like 2H^2H).

📐Formulae

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

T(K)=t(∘C)+273T(K) = t(^{\circ}C) + 273

ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T

Lt=L0(1+αΔT)L_t = L_0(1 + \alpha \Delta T)

γ=3α\gamma = 3\alpha

Density (ρ)=Mass (m)Volume (V)\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}

💡Examples

Problem 1:

An iron rod of length 5 m5\text{ m} is heated from 20∘C20^{\circ}C to 120∘C120^{\circ}C. Calculate the increase in length if the coefficient of linear expansion α=1.2×10−5 ∘C−1\alpha = 1.2 \times 10^{-5} \text{ }^{\circ}C^{-1}.

Solution:

Given: L0=5 mL_0 = 5\text{ m}, ΔT=120−20=100∘C\Delta T = 120 - 20 = 100^{\circ}C. Using ΔL=L0αΔT=5×(1.2×10−5)×100=0.006 m\Delta L = L_0 \alpha \Delta T = 5 \times (1.2 \times 10^{-5}) \times 100 = 0.006\text{ m} or 6 mm6\text{ mm}.

Explanation:

Thermal expansion causes the molecules to vibrate more vigorously, increasing the average distance between them, which results in an increase in the macroscopic length of the rod.

Problem 2:

Convert the normal human body temperature 37∘C37^{\circ}C into Fahrenheit (∘F^{\circ}F) and Kelvin (KK).

Solution:

For Fahrenheit: 375=F−329  ⟹  7.4×9=F−32  ⟹  66.6+32=98.6∘F\frac{37}{5} = \frac{F - 32}{9} \implies 7.4 \times 9 = F - 32 \implies 66.6 + 32 = 98.6^{\circ}F. For Kelvin: T=37+273=310 KT = 37 + 273 = 310\text{ K}.

Explanation:

This demonstrates the use of standard conversion scales between Celsius, Fahrenheit, and the absolute Kelvin scale.

Problem 3:

Why do water pipes often burst in cold countries during winter when the temperature falls below 0∘C0^{\circ}C?

Solution:

Due to the anomalous expansion of water, when the temperature falls from 4∘C4^{\circ}C to 0∘C0^{\circ}C, water expands. Upon freezing into ice at 0∘C0^{\circ}C, its volume increases further by about 9%9\%.

Explanation:

The expansion of water as it turns to ice exerts tremendous pressure on the walls of the pipes, causing them to burst if they are not strong enough to withstand the force.