Water falls from a height 500 m. What is the rise in the temperature of water at the bottom, if the whole energy remains in water?(Specific heat of water = 4300 J/kg °C)
Energy is the ability to do work. It exists in various forms, like kinetic, potential, thermal and chemical. Energy powers everything, from machines to living beings. It cannot be created or destroyed, only transformed. Renewable sources, like solar and wind, are vital for a sustainable future, reducing dependence on fossil fuels.
Class 11 Physics Chapter 10 Thermal Properties of Matter focuses on the concepts of heat, temperature, and their effects on matter. It covers thermal expansion, specific heat capacity, calorimetry, and heat transfer methods such as conduction, convection and radiation. Understanding these concepts is crucial for CBSE Exam 2024-25 preparation and real-world applications in science and engineering.
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To solve this we analyze this situation by applying the concept of energy conversion as follows;
Step 1: Potential Energy converted to heat To indicate that whenever water drops from a height of 500 m, its potential energy is turned into heat. The PE per unit mass is thus given as: PE = mgh where: – m is the mass of water in kilograms
– g = 9.8 m/s² is the acceleration due to gravity,
– h = 500 m is the height.
So:
PE = m ⋅ 9.8 ⋅ 500
Step 2: Energy to temperature rise
The heat produced is utilized to raise the temperature of water. The heat equation is:
Q = m ⋅ c ⋅ ΔT
where:
– Q is the heat energy,
– c = 4300 J/kg°C is the specific heat of water,
– ΔT is the rise in temperature.
Q = PE:
m ⋅ 9.8 ⋅ 500 = m ⋅ 4300 ⋅ ΔT
Step 3: Simplify and solve for ΔT
Cancel m from both sides:
9.8 ⋅ 500 = 4300 ⋅ ΔT
Solve for ΔT: ΔT = (9.8 ⋅ 500) / 4300 = 4900 / 4300 ≈ 1.16°C
Final Answer:
The temperature of the water at the bottom has increased by 1.16°C.
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