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ad-work [718]
3 years ago
8

A low-temperature physicist wishes to publish his experimental result that the heat capacity of a nonmagnetic dielectric materia

l between 0.05K and 0.5K varies as
A√ T + BT^3

As editor of the journaL should you accept the paper for publication?
Engineering
1 answer:
tatiyna3 years ago
6 0

Answer:

Yes, this because the specific heat does not violate the third law of thermodynamics.

Explanation:

The third law of thermodynamic is usually used for a closed system to relate the thermodynamic properties of the system at equilibrium conditions. For this law, a body that exists at a temperature of 0 K will cease to move or stop moving. It can be inferred that heat capacity at (T = 0 K) is equivalent to 0. Therefore, the equation is valid for the given temperature range.  

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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
4. Lockout/tagout (LOTO) is a safety procedure that ensures dangerous machines are properly shut off and not started up again pr
klemol [59]

Answer:true

Explanation:

5 0
3 years ago
fdkgdsvdgvdfgvsdcvbfbfdbvfdbsdvbesgvdslgfkrledmgoskflodjgloerjgvoljgegjp;erorf;wgp;kiaers;ogjo;rhgerjfrejgfdlhodjglodjheihtgo;rg
RSB [31]

Answer:

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pw]oe;iLH[U

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Explanation:

3 0
3 years ago
use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square
Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25  

 = 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000  

 = 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives  

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh  

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as  

E = 968.75* 103 / 30000  

E = 32.291 KBtu/sf.

8 0
4 years ago
find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters an
Anna35 [415]

Answer:

The volume for this is 29.7

Explanation:

Trust me on this I'm an expert

8 0
3 years ago
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