   Chapter 12, Problem 70AP

Chapter
Section
Textbook Problem

Every second at Niagara Falls, approximately 5.00 × 103 m3 of water falls a distance of 50.0 m. What is the increase in entropy per second due to the falling water? Assume the mass of the surroundings is so great that, its temperature and that of the water stay nearly constant at 20.0°C. Also assume a negligible amount of water evaporates.

To determine
The increase in entropy per second due to the falling water.

Explanation

Given Info:

The water falls to the waterfalls is 5.0×103m3 .

The height at which the water falls is 50m .

The temperature of the water falls is 20.0°C .

The potential energy of the water which is falling from the height is transformed to the thermal energy when the falling water hits the bottom point of the water falls.

Thus, the rate at which the thermal energy production is,

ΔQΔt=(ΔmΔt)gh (I)

• g is the acceleration due to gravity
• h is the height at which the water falls

Since, Δm=ρΔV

Equation (I) gives,

ΔQΔt=ρ(ΔVΔt)gh

Formula to calculate the rate of entropy production is,

ΔSΔt=ΔQ/ΔtT=ρ(ΔV/Δt)ghT

• ρ is the density of the water
• T is the temperature of the water

Substitute 20.0°C for T, 103kg/m3 for ρ , 5×103m3/s for ΔV/Δt , 9

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