krit.club logo

Earth as a System: Energy, Matter and Life - Uneven Heating of the Earth

Grade 9CBSE

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

🔑Concepts

•

The primary source of energy for Earth is Solar Radiation (Insolation). Due to the Earth's spherical shape, the angle of incidence of solar rays varies with latitude, leading to maximum heating at the equator and minimum heating at the poles.

•

The tilt of the Earth's axis at an angle of 23.5∘23.5^\circ to the vertical (or 66.5∘66.5^\circ to its orbital plane) causes the change of seasons and variation in day length, contributing to uneven heating.

•

Specific Heat Capacity: Land surfaces have a lower specific heat capacity than water (cland<cwaterc_{land} < c_{water}). Consequently, land heats up and cools down much faster than the oceans, causing differential heating between the lithosphere and hydrosphere.

•

Albedo effect: Different surfaces reflect different amounts of solar energy. The Albedo α\alpha is the ratio of reflected radiation to incident radiation. Fresh snow has a high albedo (approx. 0.80.8 to 0.90.9), while dark soil has a low albedo (approx. 0.10.1).

•

Pressure Belts: Uneven heating creates pressure gradients. Warm air at the equator rises, creating low pressure (LL), while cold air at the poles sinks, creating high pressure (HH). This movement forms global wind patterns through convection currents.

•

The Coriolis Effect: Due to Earth's rotation, winds are deflected to the right in the Northern Hemisphere and to the left in the Southern Hemisphere, modifying the path of heat transfer across the globe.

📐Formulae

Q=mcΔTQ = mc\Delta T

Albedo (α)=Reflected RadiationTotal Incident Radiation\text{Albedo } (\alpha) = \frac{\text{Reflected Radiation}}{\text{Total Incident Radiation}}

I=S⋅sin⁡(θ)I = S \cdot \sin(\theta)

Net Radiation=(1−α)SWin+LWin−LWout\text{Net Radiation} = (1 - \alpha)SW_{in} + LW_{in} - LW_{out}

💡Examples

Problem 1:

During a sunny day, 1000 J1000 \text{ J} of heat energy is absorbed by 1 kg1 \text{ kg} of dry soil (cs=800 J/kg∘Cc_s = 800 \text{ J/kg}^\circ\text{C}) and 1 kg1 \text{ kg} of water (cw=4184 J/kg∘Cc_w = 4184 \text{ J/kg}^\circ\text{C}). Calculate the rise in temperature for both and explain which one heats up faster.

Solution:

Using the formula ΔT=Qmc\Delta T = \frac{Q}{mc}:

For Soil: ΔTsoil=10001×800=1.25∘C\Delta T_{soil} = \frac{1000}{1 \times 800} = 1.25^\circ\text{C}

For Water: ΔTwater=10001×4184≈0.239∘C\Delta T_{water} = \frac{1000}{1 \times 4184} \approx 0.239^\circ\text{C}

Explanation:

Since 1.25∘C>0.239∘C1.25^\circ\text{C} > 0.239^\circ\text{C}, the soil (land) heats up significantly faster than the water. This temperature difference is the fundamental cause of sea breezes during the day and land breezes at night.

Problem 2:

Calculate the Albedo of a glacier surface if it receives 400 W/m2400 \text{ W/m}^2 of solar radiation and reflects back 320 W/m2320 \text{ W/m}^2.

Solution:

Using the Albedo formula: α=320400\alpha = \frac{320}{400} α=0.8\alpha = 0.8

Explanation:

The albedo is 0.80.8 or 80%80\%. High albedo values mean the surface reflects most of the energy, which is why polar regions remain cold even when receiving sunlight during the summer months.