/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 At the top a mountain the temper... [FREE SOLUTION] | 91Ó°ÊÓ

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At the top a mountain the temperature is \(-5^{\circ} \mathrm{C}\) and \(\mathrm{a}\) mercury barometer reads \(566 \mathrm{~mm}\), whereas the reading at the foot of the mountain is \(749 \mathrm{~mm}\). Assuming a temperature lapse rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\) and \(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\), calculate the height of the mountain. (Neglect thermal expansion of mercury.)

Short Answer

Expert verified
The height of the mountain is approximately 2199.6 meters.

Step by step solution

01

Convert Temperature to Kelvin

Convert temperature at the top of the mountain from Celsius to Kelvin. Add 273.15 to -5°C.\[-5 + 273.15 = 268.15 \text{ K}\]
02

Use the Barometric Formula

The height difference (\Delta h) can be found by the barometric formula:\[P_2 = P_1 \times \text{exp}\bigg(-\frac{g \times \Delta h}{R \times T}\bigg)\]Here,\ P_2 = 566 \text{ mmHg (Pressure at the top)}\ \P_1 = 749 \text{ mmHg (Pressure at the bottom)}\ g = 9.81 \text{ m/s}^2\ \ R = 287 \text{ J/kg/K}\ \ T = 268.15 \text{ K (average temperature)}\
03

Set Up the Equation

Rearrange the barometric formula to solve for \Delta h:\[\Delta h = -\bigg(\frac{R \times T}{g}\bigg) \times \text{ln}\bigg(\frac{P_2}{P_1}\bigg)\]Substitute the known values into the equation:\[\Delta h = -\bigg(\frac{287 \times 268.15}{9.81}\bigg) \times \text{ln}\bigg(\frac{566}{749}\bigg)\]
04

Calculate the Natural Logarithm

Calculate the natural logarithm:\[\text{ln}\bigg(\frac{566}{749}\bigg) = \text{ln}(0.755) = -0.280\]
05

Compute the Height Difference

Compute the height difference using the values:\[\Delta h = -\bigg(\frac{287 \times 268.15}{9.81}\bigg) \times (-0.280) = \approx 2199.6 \text{ meters}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Lapse Rate
The temperature lapse rate is a key factor when calculating altitude based on pressure and temperature differences. It represents the rate at which temperature decreases with an increase in altitude. In many atmospheric conditions, this rate is standardized at 0.0065 K/m, meaning for every meter you ascend, the temperature drops by 0.0065 Kelvin.
For example, if you start at sea level with a temperature of 15°C and climb 1000 meters, you would experience a decrease in temperature by approximately 6.5 K, ending up around 8.5°C. Understanding this rate helps us accurately determine the changes in atmospheric pressure and, consequently, in altitude.
Pressure Difference
Pressure difference is a crucial component in altitude calculations. In the given problem, the pressures at the top and bottom of the mountain were 566 mmHg and 749 mmHg, respectively. Atmospheric pressure decreases with altitude due to the thinning of air.
By measuring the pressure difference and utilizing the barometric formula, we can infer the height difference between two points. It's important to convert pressure units consistently. For instance, 1 mmHg (millimeter of mercury) is equivalent to approximately 133.322 Pa (Pascal). The relative decrease in pressure as altitude increases is essential for determining height accurately.
Altitude Calculation
Altitude calculation involves combining temperature and pressure data to estimate height differences. Using the barometric formula \[P_2 = P_1 \times \text{exp}\bigg(-\frac{g \times \bigtriangleup h}{R \times T}\bigg)\], we link atmospheric pressure changes to altitude.
Here, \[P_2\] and \[P_1\] are the pressures at two altitudes, \[g\] is the gravitational constant (9.81 m/s²), \[R\] is the specific gas constant for dry air (287 J/kg/K), and \[T\] is the average temperature. Converting pressures to the same units and solving for \[\bigtriangleup h\] using the logarithmic form of the equation helps us determine the height of the mountain, which in this case is approximately 2199.6 meters.
Fluid Mechanics
Fluid mechanics principles are embedded in the barometric formula calculation. The atmosphere can be considered a fluid, and we analyze its behavior under varying conditions of pressure, temperature, and density. Fluid dynamics and statics govern how atmospheric pressure changes with altitude.
Understanding fluid mechanics in this context allows us to apply principles such as hydrostatic equilibrium, where the weight of the atmosphere at a given height balances the pressure forces. This balance is what leads to the exponential relation in the barometric formula, enabling us to link pressure differences with altitude changes accurately.

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Most popular questions from this chapter

A spherical, helium-filled balloon of diameter \(800 \mathrm{~mm}\) is to be used to carry meteorological instruments to a height of \(6000 \mathrm{~m}\) above sea level. The instruments have a mass of \(60 \mathrm{~g}\) and negligible volume, and the balloon itself has a mass of \(100 \mathrm{~g}\). Assuming that the balloon does not expand and that atmospheric temperature decreases with increasing altitude at a uniform rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\), determine the mass of helium required. Atmospheric pressure and temperature at sea level are \(15^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\) respectively; for air, \(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are \(2 \mathrm{~m}\) long and the tank contains a liquid of relative density 1.15. The centre of the aperture is \(1.5 \mathrm{~m}\) below the free surface. Calculate the net hydrostatic thrust on the plate, the moment of this thrust about the hinge and the position of the centre of pressure.

A solid cylinder, \(1 \mathrm{~m}\) diameter and \(800 \mathrm{~mm}\) high, is of uniform relative density \(0.85\) and floats with its axis vertical in still water. Calculate the periodic time of small angular oscillations about a horizontal axis.

A buoy, floating in sea-water of density \(1025 \mathrm{~kg} \cdot \mathrm{m}^{-3}\), is conical in shape with a diameter across the top of \(1.2 \mathrm{~m}\) and a vertex angle of \(60^{\circ}\). Its mass is \(300 \mathrm{~kg}\) and its centre of gravity is \(750 \mathrm{~mm}\) from the vertex. A flashing beacon is to be fitted to the top of the buoy. If this unit is of mass \(55 \mathrm{~kg}\) what is the maximum height of its centre of gravity above the top of the buoy if the whole assembly is not be unstable? (The centroid of a cone of height \(b\) is at \(3 h / 4\) from the vertex.)

An open-topped tank, in the form of a cube of \(900 \mathrm{~mm}\) side, has a mass of \(340 \mathrm{~kg}\). It contains \(0.405 \mathrm{~m}^{3}\) of oil of relative density \(0.85\) and is accelerated uniformly up a long slope at arctan (1/3) to the horizontal. The base of the tank remains parallel to the slope, and the side faces are parallel to the direction of motion. Neglecting the thickness of the walls of the tank, estimate the net force (parallel of the slope) accelerating the tank if the oil is just on the point of spilling.

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