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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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3 years ago
A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
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Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

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\frac{3+r}{r}=3.095

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Thus Electric field is zero at some distance r=1.43 cm right of q_2

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Answer:

Given,

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T = 393;;K

Convert to Celcius;

T = (393-273) degrees

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vf = 0.001060 m³/kg

vg = 0.89133 m³/kg

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Specific volume is given as;

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