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4 years ago

The laws of thermodynamics state that, in a heat engine, _____.

Physics

2 answers:

Daniel [21]4 years ago

5 0

Answer: The correct option is "all the heat energy from a source cannot be converted to mechanical energy"

Explanation:

Heat engine is a system which converts heat energy to the mechanical energy. During this process, a lot of heat energy is lost to surrounding.

Heat engine is a device that uses heat to do work. It is impossible to extract an amount of heat from the reservoir at high temperature and use this heat to do some work.

Some heat is rejected to the reservoir at low temperature.

Therefore, the laws of thermodynamics state that, in a heat engine, all the heat energy from a source cannot be converted to mechanical energy.

olchik [2.2K]4 years ago

4 0

<span>There are four laws of thermodynamic which define and characterize the thermodynamic system at thermal equilibrium.
The laws of thermodynamics state that, in a heat engine, </span>all the heat energy from a source cannot be converted to mechanical energy.

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Water is made of two hydrogen atoms and one oxygen atom bonded together. Julia is describing how water undergoes a physical chan

FrozenT [24]

Everything that Julia said is correct EXCEPT the words
"the Oxygen and Hydrogen separate". They do not.

The "particles" that she mentioned are water molecules,
each one composed of two Hydrogen atoms and one Oxygen
atom bonded together. THOSE are the particles that shift to be
farther apart, but the Hydrogen and Oxygen in each one remain
bonded together.

Julia gets 30 ounces of merit badges, out of a possible quart.

6 0

3 years ago

If a car used 260,000 W of power to complete a race in 15 s, how much work did the car do?

suter [353]

Answer: 3.9 MW

Explanation:

1 W = 1 J/s

260000 J/s (15 s) = 3,900,000 = 3.9 MW

3 0

3 years ago

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the

My name is Ann [436]

Answer:

2.82 s

Explanation:

The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 is the starting position, 2.3 m in this case.

Vy0 is the starting speed, 13 m/s.

a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.

Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2

It will reach the ground when Y(t) = 0

0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2

-4.9 * t^2 + 13 * t + 2.3 = 0

Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

7 0

3 years ago

If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?

natulia [17]

If a galaxy is located 200 million light years from Earth, you can conclude that t<span>he light will take 200 million years to reach Earth. </span>

8 0

3 years ago

Read 2 more answers

A high-temperature, gas-cooled nuclear reactor consists of a composite, cylindrical wall for which a thorium fuel element (kth =

WARRIOR [948]

Answer:

a) T_1 = 938 K , T_2 = 931 K

b) To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above:

q = 3*10^8 W/m^3

Explanation:

Given:

- See the attachment for the figure for this question.

- Melting point of Thorium T_th = 2000 K

- Melting point of Thorium T_g = 2300 K

Find:

a) If the thermal energy is uniformly generated in the fuel element at a rate q = 10^8 W/m^3 then what are the temperatures T_1 and T_2 at the inner and outer surfaces, respectively, of the fuel element?

b) Compute and plot the temperature distribution in the composite wall for selected values of q. What is the maximum allowable value of q.

Solution:

part a)

- The outer surface temperature of the fuel, T_2, may be determined from the rate equation:

q*A_th = T_2 - T_inf / R'_total

Where,

A_th: Area of the thorium section

T_inf: The temperature of coolant = 600 K

R'_total: The resistance per unit length.

- Calculate the resistance per unit length R' from thorium surface to coolant:

R'_total = Ln(r_3/r_2) / 2*pi*k_g + 1 / 2*pi*r_3*h

Plug in values:

R'_total = Ln(14/11) / 2*pi*3 + 1 / 2*pi*0.014*2000

R'_total = 0.0185 mK / W

- And the heat rate per unit length may be determined by applying an energy balance to a control surface about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that:

q' = q*A_th = q*pi*(r_2^2 - r_1^2)

q' = 10^8*pi*(0.011^2 - 0.008^2) = 17,907 W / m

Hence,

T_2 = q' * R'_total + T_inf

T_2 = 17,907*0.0185 + 600

T_2 = 931 K

- With zero heat flux at the inner surface of the fuel element, We will apply the derived results for boundary conditions as follows:

T_1 = T_2 + (q*r_2^2/4*k_th)*( 1 - (r_1/r_2)^2) - (q*r_1^2/2*k_th)*Ln(r_2/r_1)

Plug values in:

T_1 = 931+(10^8*0.011^2/4*57)*( 1 - (.8/1.1)^2) - (10^8*0.008^2/2*57)*Ln(1.1/.8)

T_1 = 931 + 25 - 18 = 938 K

part b)

The temperature distributions may be obtained by using the IHT model for one-dimensional, steady state conduction in a hollow tube. For the fuel element (q > 0), an adiabatic surface condition is prescribed at r_1 while heat transfer from the outer surface at r_2 to the coolant is governed by the thermal resistance:

R"_total = 2*pi*r_2*R'_total

R"_total = 2*pi*0.011*0.0185 = 0.00128 m^2K/W

- For the graphite ( q = 0 ), the value of T_2 obtained from the foregoing solution is prescribed as an inner boundary condition at r_2, while a convection condition is prescribed at the outer surface (r_3).

- For 5*10^8 < q and q > 5*10^8, the distributions are given in attachment.

The graphs obtained:

- The comparatively large value of k_t yields small temperature variations across the fuel element, while the small value of k_g results in large temperature variations across the graphite.

Operation at q = 5*10^8 W/^3 is clearly unacceptable, since the melting points of thorium and graphite are exceeded and approached, respectively. To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above:

q = 3*10^8 W/m^3

6 0

3 years ago