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3 years ago

If the change in enthalpy is -5074.3 kj, how much work is done during the combustion?

1 answer:

6 0

<span>Answer: I guess there is is a typo in your question both change in enthalpy and change in internal energy for the combustion are negative: â†H = -5074 kJ â†U = - 5084.2 kJ Work done to surrounding in a process is given by the integral: W = â« P dV from initial to final volume for a constant pressure process the integral simplifies to: W = P â™ â« dV = Pâ™â†V Enthalpy is defined as: H = U + Pâ™V So the change in enthalpy and the change in internal energy are related as: â†H = â†U + â†(Pâ™V) for a constant pressure process: â†H = â†U + Pâ™â†V Hence, â†H = â†U + W => W = â†H - â†U = -5074 kJ - (-5084.2) kJ = 10.2 kJ</span>

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A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

3 0

2 years ago

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago

What is the charge of the negative sphere

tatyana61 [14]

Answer:

electrons

Explanation:

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3 years ago

A major feature of the solar system is that as planets get farther away from the sun they, _________ a. Are warmer b. Have fewer

defon

D.
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2 years ago

A cannonball is shot up in the air with a vertical speed of 24 miles what is the cannon balls vertical speed just before it hits

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Answer:

I think it will back towards the earth because earth gravitional field will attract to Wards

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2 years ago