When a substance goes directly from a gaseous state to a solid state as dry ice does

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t

Tcecarenko [31]

Answer:

The amount of heat transfer is 21,000J .

Explanation:

The equation form of thermodynamics is,

ΔQ=ΔU+W

Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.

Substitute 0 J for W and 0 J for ΔU

ΔQ = 0J+0J

ΔQ = 0J

The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change

The heat transfer is,

ΔQ=Q (in ) −Q (out )

Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)

ΔQ=(19000J+2000J)−(0J)

=21,000J

Thus, the amount of heat transfer is 21,000J .

8 0

3 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

(c) $E=2.24*10^{4} N/C$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz

gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa

Tt is stagnation temperature = ?

T is the static temperature = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 = Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0

2 years ago

a 15 kg tv sots on a shelf at a height of 0.3 m. how much gravitational potential energy is added to the television when it is l

nata0808 [166]

Gravitational potential energy can be described as m*g*h (mass times gravity times height).

Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.

After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.

To find how much energy was added, we subtract final energy from initial energy:

147 J - 44.1 J = 102.9 Joules.

6 0

3 years ago

I WILL MARK YOU THE BRAINLIEST NO LINKS

alexandr402 [8]

The strength of the force of friction depends on two factors; the type of surfaces involves and the force that the surfaces are being push together

3 0

3 years ago