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katovenus [111]

4 years ago

9

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

evation where the pressure is 0.77 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?

1 answer:

Irina-Kira [14]4 years ago

4 0

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

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alexandr1967 [171]

Answer:

The value is $h = 11930 \ m$

Explanation:

From the question we are told that

The temperature is $T = 300 \ K$

Generally the root mean square speed of the oxygen molecules is mathematically represented as

$v = \sqrt{\frac{3 * R * T }{M} } = \sqrt{ 2 * g * h }$

Here R is the gas constant with a value $R = 8.314 \ J\cdot K^{-1} \cdot \ mol^{-1}$

M is the molar mass of oxygen molecule with value $M = 0.032 \ kg /mol$

So

$\frac{3 * 8.314 * 300 }{0.032} = 2 * 9.8 * h$

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Determine the volume displaced and then calculate the density of this 54 g sample of brass.

inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

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Buoyant force exerted by a liquid on an object with $V_{imm}$ of it's volume immersed is :

$F_{B}=V_{imm}*d_{l}*g$

where ,

$F_{B}$ is the buoyant force
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$g$ is the acceleration due to gravity

Thus at equilibrium:

$m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}$

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3 years ago

A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur

Anna35 [415]

Answer:

$\dot{W_{H} } = 4244.48 Btu/h$

Explanation:

Temperature of the house, $T_{H} = 70^{0} F$

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$\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\$ ...............(*)

Where the heat losses from the house, $\dot{Q_{H} } = 75,000 Btu/h$

Substituting these values into * above

$\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h$

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3 years ago