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A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate

yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

$W=Pm\DeltaV$

$W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}$

$W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})$

Where,R = gas constant

T = temperature

P = pressure

$P_{atm}$ =Atmosphere pressure

m = mass

Put the value into the formula

$W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)$

$W=213\ kJ$

Hence, The work done by the steam is 213 kJ.

6 0

3 years ago

An alternator provides 20 A of peak current at 100 V peak AC to a resistive electric heater. How much heating power is delivered

Liono4ka [1.6K]

-- The resistance of the heater is (volts/current) = 5 ohms

-- The heating (RMS) value of a sinusoidal AC is V(peak)/√2 . For this particular alternator, V(peak)=100V, so the heating (RMS) equivalent is 70.71 V.

-- The heating power delivered to the electric heater is (E²/R).

Power = (100/√2)² / 5

Power = 5,000 / 5

Power = 1,000 watts

6 0

3 years ago

Which of the following could possibly be predicted? Earthquakes Landslides Foreshocks Volcanic eruptions

NeTakaya

Answer:

Volcanic Eruptions

Explanation:

The volcano can start showing signs that it may be about to explode.

6 0

3 years ago

How much heat is needed to boil 120 kg of water ?

nekit [7.7K]

Q = mc∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

Hope I helped!! xx

7 0

3 years ago

You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w

solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

= 0.025×< -2, 0, -7 >

= < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

= 0.025×< 4, 4, -3 >

= < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact = < -0.05, 0, 0.175 > + < 0.1, 0.1, -0.075>

= < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity = v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

= < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0

3 years ago