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3 years ago

8

A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

what is the new volume of the balloon in the hot room?

2 answers:

grin007 [14]3 years ago

8 0

I think the answer is 85 degrees

Anestetic [448]3 years ago

6 0

this would be equal to 3.68 L

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Answer:

Time = 12.43 years

Mass = 4.71.43kg

Explanation:

Given the following :

Distance = 8.7 light years

Speed = 70% the speed of light

Diameter = 6m

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Density = 1 hydrogen atom/m^3

Mass of proton(Me) = 1.673x 10^-27

A.) time it takes your spaceship to reach Sirius :

From the relation: Speed = (distance / time)

Time = distance / speed

Time = (70/100) × 1 light year

Distance = 8.7 light years

Time = 8.7 / 0.7 = 12.4285 years

Time = 12.43 years

B.) Mass of inter-stellar gas that collides with the spaceship can be calcuted by finding the product of the surface area of the cylindrical space ship and the mass of proton.

That is ;

surface area * mass of proton

Surface area of a cylinder = 2πrh + πr^2×Me

= 2πrh + Me×πr^2)

=( 2 × 22/7 × 3 × 25) + (22/7 × 3^2 × 1.673 * 10^-27)

= 2× 235.714 + 28.285) × 1.673 * 10^-27

= 471.428 + 47.31 * 10^-27

= 471.428 + 4.73 × 10^-26

= 4.71.43kg approximately

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The BERMAD model 400Y-5DC is an elastomeric, hydraulic line pressure operated deluge valve, designed specifically for advanced fire protection systems and the latest industry standards. The 400Y-5DC is activated by a hydraulically operated relay valve, through which opening and closing of the valve can be controlled either with a remote hydraulic command or with a wet pilot line with closed fusible plugs. An integral pressure reducing pilot valve ensures a precise, stable, pre-set downstream water pressure. The optional valve position indicator can include a limit switch suitable for Fire & Gas monitoring systems. The 400Y-5DC is ideal for systems that combine a remote wet pilot line with a high pressure water supply.

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An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

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Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

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However, from the evaporator, the heat transfer Q can be determined by using the formula:

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where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

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LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

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