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Ira Lisetskai [31]
2 years ago
7

A chart labeled table A : effect of height on temperature with initial temperature as 25 degrees Celsius, mass w is 1.0 kilogram

s and mass c is 5.0 kilograms. The chart is 5 columns and 5 rows. The first column is labeled h in meters with entries 100, 200, 500, 1000. The second column is labeled temperature final in degrees celcius with entries 26.17, 27.34, 30.86, 36.72. The third column is labeled change in temperature in degrees celcius with entries 1.17, 2.34, 5.86, 11.72. The fourth column is labeled gravitational potential energy in kilojoules with no entries. The last column is labeled delta H in kilojoules with no entries.
Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tenth.
Physics
2 answers:
sweet [91]2 years ago
8 0

Answer: (1) 4.9 kj

(2) 9.8 kj

(3) 24.5 kj

(4) 49.0 kj

Explanation:

Let's label each cylinder from 1-4

Note that,

Gravitational Potential Energy (PEg) = m × g × h

Where m is mass

g is acceleration due to gravity

h is height

So, calculate the PEg of every cylinder

PEg Cylinder 1 = m × g × h

= 5 × 100 × 9.8 = 4900 J

Which can be converted to, 4900 ÷ 1000 = 4.9 kj  because from the question above, it is in kilojoules (kj)

Cylinder 2 = m × g × h

= 5 × 200 × 9.8 = 9800 J

Which can be converted to, 9800 ÷ 1000 = 9.8 kj

Cylinder 3 = m × g × h

= 5 × 500 × 9.8 = 24,500 J

Which can be converted to,  24500 ÷ 1000 = 24.5 kj

Cylinder 4 = m × g × h

= 5 × 1000 × 9.8 = 49,000 J

Which can be converted to,  49000 ÷ 1000 = 49.0 kj

Thus, the different heat generated  by each cylinder at heights 100 m, 200 m, 500 m and 1000 m (to the nearest tenth) are 4.9 kj, 9.8 kj, 24.5 kj and 49.0 kj respectively.

o-na [289]2 years ago
7 0

Answer:

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ

Explanation:

edge2020

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In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40
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Answer:

Bulk modulus = 1.35 × 10^{10} Pa

Explanation:

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density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

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we know Bulk Modulus = v^2*\rho   ...............

here \rho is density i.e 1400 kg/m³

and v is =  frequency × wavelength

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A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
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Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

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 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

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