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KiRa [710]
4 years ago
11

What are units for heating capacity? Mark all that apply: a)- Tons b)- kJ/kg c)- kW d)- Btu

Engineering
1 answer:
svet-max [94.6K]4 years ago
3 0

Answer:

(b) kJ/kg

Explanation:

The ratio of amount of energy required to change the temperature of the substance by certain magnitude and this magnitude of temperature change is known as heat capacity of the substance.

The expression for Heat capacity is:

C=E/ΔT

Where,

C is the Heat capacity

E is the energy absorbed/released

ΔT is the change in temperature

<u>The SI unit of heat capacity is J/K.</u>

(a) Tons represents the unit of mass (1000 kg)

(c) kW represents the unit of power (1000 W)

(d) Btu represents the unit of heat (1055 J)

<u>The units from the options that can be a unit of heat capacity is (b) kJ/kg.</u>

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One kilogram of water fills a 150 L rigid container at an initial pressure of 2MPa. The container is cooled to 40 oC. Find the i
Tatiana [17]

Answer:

The initial temperature is 649 K (376 °C).

The final pressure is 0.965 MPa

Explanation:

From the ideal gas equation

PV = nRT

P is the initial pressure of water = 2 MPa = 2×10^6 Pa

V is intial volume = 150 L = 150/1000 = 0.15 m^3

n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol

R is gas constant = 8.314 m^3.Pa/mol.K

T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C

From pressure law,

P1/T1 = P2/T2

P2 (final pressure) = P1T2/T1

T2 (final temperature) = 40 °C = 40 + 273 = 313 K

P1 (initial pressure) = 2 MPa

T1 (initial temperature) = 649 K

P2 = 2 × 313/649 = 0.965 MPa

5 0
4 years ago
Read 2 more answers
Water flows near a flat surface and some measurements of the water velocity, u, parallel to the surface, at different heights y
BabaBlast [244]

Answer:

a) since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b) The velocity according to the equation at y=0 is equal to 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary, this value is wrong hence the equation is NOT CORRECT

Explanation:

Given that;

range ⇒ 0 < y < 0

1ft is given by the equation u = 0.81 + 9.2y + (4.1 × 10³y³)

so u=velocity of water at different layers

y= height of the layer

a)

consider BG system of units

u(ft/s) = 0.81 + 9.2y + (4.1 × 10³y³)

and consider y=0.05 ft

u = 0.81 + 9.2(0.5) + (4.1 × 10³(0.5³)

u = 0.81 + 0.46 + 0.5125

u = 1.7825 ft/s lets say this is equation 1

now consider the SI system units

u(m/s) = 0.81 + 9.2y + (4.1 × 10³y³)

also consider y=0.05ft

1ft = 3.048×10⁻¹ (from conversion table)

so 0.05ft = 0.01524m

we substitute

u(m/s) = 0.81 + 9.2(0.01524m) + (4.1 × 10³(0.01524m)³)

u = 0.81 + 0.1402 + 1.4512×10⁻²

u = 0.9647 m/s

1m/s = 3.281 ft per seconds ( conversion table)

so

0.9647 m/s = 0.9647(3.281)

u = 3.165 ft/s lets say this is equation 2

now since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b)

we know that the velocity of water at the surface contact is zero

u=0

so from the equation

u = 0.81 + 9.2y + (4.1 × 10³y³)

at y = 0

u = 0.81 + 9.2(0) + (4.1 × 10³(0)³)

u = 0.81 ft/s

The velocity according to the above equation at y=0 is 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary this value is wrong hence the equation is NOT CORRECT

5 0
4 years ago
..........................
kow [346]

Answer:

i wish i knew what you meant by that

7 0
3 years ago
Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai
Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

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4 years ago
Pls pls pls pls plsss someone help pls
snow_tiger [21]

Answer:

I would like to help can you take a picture without the bot telling you something. Maybe then I can figure it out.

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