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3 years ago

URGENT PLEASE HELP!!!

Engineering

1 answer:

diamong [38]3 years ago

4 0

Answer:

Q = 292.6 kJ

Explanation:

First, let's make sure all the values are in appropriate units.

C (specific heat of water) = 4.18 J/g °C

m (mass of water) = 2000 g

∆T = 35 °C

The equation for the needed amount of energy is:

Q = m • C • ∆T

So, we simply multiply the given values:

Q = 2000 • 4.18 • 35

Q = 292600 J

If we want the answer in kilojoules (kJ), we simply divide the value in joules with 1000:

Q = 292600 / 1000 = 292.6 kJ

I hope this helps.

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svet-max [94.6K]

Answer:

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3 years ago

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g

Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$ $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³

solution

we get here N that is

N = $\frac{N_A \times \rho}{A}$ ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$ .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0

3 years ago

Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12

Katarina [22]

Answer:

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

$- \dot W_{out} + \dot m \cdot (h_{in} - h_{out}) = 0$

The turbine power output is:

$\dot W_{out} = \dot m\cdot (h_{in}-h_{out})$

The volumetric flow is:

$\dot V = \frac{\pi}{4} \cdot \left( \frac{2}{12}\,ft \right)^{2}\cdot (620\,\frac{ft}{s} )$

$\dot V \approx 13.526\,\frac{ft^{3}}{s}$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,\frac{ft^{3}}{lbm}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu}$

$\dot m = \frac{13.526\,\frac{ft^{3}}{s} }{1.33490\,\frac{ft^{3}}{lbm} }$

$\dot m = 10.133\,\frac{lbm}{s}$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,\frac{BTU}{lbm}$

State 2 (Saturated Vapor)

$h = 1146.1\,\frac{BTU}{lbm}$

The turbine power output is:

$\dot W_{out} = (10.133\,\frac{lbm}{s} )\cdot (1479.1\,\frac{BTU}{lbm}-1146.1\,\frac{BTU}{lbm})$

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

6 0

3 years ago

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Sindrei [870]

Answer:

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Explanation:

4 0

3 years ago

What is the thermal efficiency of a gas power cycle using thermal energy reservoirs at 627°C and 60°C? The thermal efficiency is

Vesna [10]

Answer

63 %

Explanation:

It is given that the reservoirs is at the temperature of 627°C and 27°C

So lower temperature that is $T_L$ = 60°C=273+60=333 K

And the higher temperature that is $T_H$ = 627°C =273+627=900 K

We know that the thermal efficiency of thermal reservoir = 1- $\frac{T_L}{T_H}$ = $1-\frac{333}{900}$ = 0.63 =63 %

So the efficiency of the reservoir is 63%

3 0

3 years ago