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Ray Of Light [21]
3 years ago
13

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth? The bulk modulus of water is 2.3 × 109 Pa.
Physics
1 answer:
dangina [55]3 years ago
8 0

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

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3 years ago
An air bubble of volume 20 cm³ is at the bottom of a lake 40 m deep, where the temperature is 4.0°C. The bubble rises to the sur
soldi70 [24.7K]

Answer:

100 cm³

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

n and R are constant, so:

P₁V₁/T₁ = P₂V₂/T₂

If we say point 1 is at 40m depth and point 2 is at the surface:

P₂ = 1.013×10⁵ Pa

T₂ = 20°C + 273.15 = 293.15 K

P₁ = ρgh + P₂

P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa

P₁ = 4.933×10⁵ Pa

T₁ = 4.0°C + 273.15 = 277.15 K

V₁ = 20 cm³

Plugging in:

(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)

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What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera
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Answer:

The amount of energy required is 31.1692 kJ .

Explanation:

The transitions of water are as follows,

(243K ice to 273K ice to 273K water to 373k water to 373K steam to 393K steam)

The following required data is,

specific heat capacity of ice= 2.1 kJ/kg

specific heat capacity of water= 4.2 kJ/kg

specific heat capacity of ice= 1.996 kJ/kg

specific latent heat of fusion=334 kJ/kg

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FORMULAS:-

temperature change=mc(T2-T1)

phase transistion = mL,

where, m=mass , c=specific heat capacity ,L = latent heat of fusion or vapourisation ,(T2-T1)= temparature change

Thus the total amount of heat is,

=mc(T4-T3) + mL + mc(T3-T2) + mL + mc(T2-T1)

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=(0.3)(2.1) + (3.34) + (4.2) +  (22.6) + (0.2)(1.996)

=31.1692 kJ.

3 0
3 years ago
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