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ruslelena [56]
3 years ago
7

A mass of 15 kg of air in a piston-cylinder device is heated from 25 o C to 77 o C by passing current through a resistance heate

r inside the cylinder. The pressure inside the cylinder is help constant at 300 kPa during this process and a heat loss of 60 kJ occurs. Calculate the electric energy supplied in kWh.
Physics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

The electrical energy supplied is 0.233 kWh

Explanation:

Given :

Mass of air m = 15 kg

Initial temperature T _{1} = 298 K

Final temperature T_{2} = 350 K

Heat loss Q _{out} = 60 kJ

From the first law of thermodynamics,

     \Delta U = W_{in} - W_{out} - Q_{out}

We know that internal energy is proportional to difference of temperature,

    W_{in} = m (T_{2} - T_{1}  ) + 60

    W_{in} = 15 \times  (350-298) + 60 }

Now we need answer into kWh so ( 1 kWh = 3600 kJ ).

So we need to multiply ( \frac{1}{3600} )

     W_{in} = [15 \times  (350-298) + 60 ] \times \frac{1}{3600} kWh

     W_{in} = 0.233 kWh

Therefore, the electrical energy supplied 0.233 kWh

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Explanation:

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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
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Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

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