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Heat Transfer in Nature - Seepage of water beneath the Earth

Grade 7CBSE

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

🔑Concepts

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Infiltration (Seepage): The process by which water on the ground surface enters the soil and moves into the ground is called infiltration. This process recharges the groundwater.

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Water Table: The upper limit of the underground water level is known as the Water Table. It varies from place to place and may even change at the same place depending on the season.

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Aquifer: Water stored between layers of hard rock below the water table is known as an aquifer. This water can be pumped out using tube wells or hand pumps.

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Heat Transfer in Soil: Heat is transferred through the soil primarily via Conduction (between soil particles) and Convection (as water seeps down through the pores).

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Percolation Rate: The speed at which water passes through the soil is called the percolation rate. It is different for different soil types (e.g., highest in sandy soil and lowest in clayey soil).

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Specific Heat of Water: Water has a high capacity to absorb heat, meaning as it seeps into the ground, it helps in regulating the Earth's internal temperature.

📐Formulae

Percolation Rate (mL/min)=Amount of Water (mL)Percolation Time (min)\text{Percolation Rate (mL/min)} = \frac{\text{Amount of Water (mL)}}{\text{Percolation Time (min)}}

Total Heat Transferred (Q)=m×c×ΔT\text{Total Heat Transferred (Q)} = m \times c \times \Delta T

💡Examples

Problem 1:

During an experiment, a student observed that it took 5050 minutes for 600 mL600 \text{ mL} of water to percolate through a soil sample. Calculate the percolation rate in mL/min\text{mL/min}.

Solution:

Given: Amount of water = 600 mL600 \text{ mL}, Percolation time = 50 min50 \text{ min}. Using the formula: Percolation rate=600 mL50 min=12 mL/min \text{Percolation rate} = \frac{600 \text{ mL}}{50 \text{ min}} = 12 \text{ mL/min}

Explanation:

The percolation rate is calculated by dividing the total volume of water by the time it took to disappear into the soil.

Problem 2:

A rain gauge recorded a total of 2500 mL2500 \text{ mL} of water. After a period of seepage, 1350 mL1350 \text{ mL} of water remained on the surface in a non-porous collection area. How much water seeped into the ground? (Use vertical subtraction)

Solution:

2500−13501150\begin{array}{r} 2500 \\ -1350 \\ \hline 1150 \end{array} The amount of water that seeped into the ground is 1150 mL1150 \text{ mL}.

Explanation:

To find the amount of water that underwent seepage (infiltration), we subtract the remaining surface water from the initial total volume of rainwater.