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kiruha [24]
3 years ago
11

The air contained in a room loses heat to the surroundings at a rate of 50 kJ/min while work is supplied to the room by computer

, TV, and lights at a rate of 1.2 kW. What is the net amount of energy change of the air in the room during a 30-min period?

Engineering
2 answers:
Mkey [24]3 years ago
8 0

Answer:

660KJ

Explanation:

Given

Let Q = Heat Loss from room = 50kj/min

Let W = Work Supplied to room = 1.2KW

1 kilowatt = 1 kilojoules per second

So, W = 1.2KJ/s

In heat and work (Sign Convention)

We know that

1. Heat takes positive sign when it is added to the system

2. Heat takes negative sign when it is removed from the system.

3. Work done is considered positive when work is done by the system

4. Work done is considered negative when work is done on the system.

From the above illustration, heat loss (Q) = -50KJ/Min

In 30 minutes time, Q = -50Kj/Min * 30 Min

Work done in 30 minutes = -1500 KJ

Also, work supplied = -1.2Kj/s

Work supplied to the system in 30 minutes = -1.2Kj/s * 30 minutes

W = -1.2 KJ/s * 30 * 60 seconds

W = -2160KJ

In thermodynamics (First Law)

Q = W + ΔU

-1500 = -2160 + ΔU

∆U = 2160 - 1500

∆U = 660KJ

Artemon [7]3 years ago
4 0

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

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What is the difference between absolute and gage pressure?
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Explanation:

Step1

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8 0
3 years ago
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an
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Answer:

volumetric flow rate = 0.0251 m^3/s

Velocity in pipe section 1 = 6.513m/s

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

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velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s

<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s

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