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RSB [31]
2 years ago
15

How much total heat energy input is needed to raise the temperature from 20°C to 100°C for 15-kg steel container filled with 17-

kg liquid? [cst = 450 J/kg.Co and cli = 4100 J/kg.Co]
Engineering
1 answer:
frez [133]2 years ago
5 0

Answer:

Total heat energy input needed is 6116kJ

Explanation:

This problem bothers heat capacity

Assuming that the steel container and the liquid have same temperatures

Given data

Initial temperature of liquid and steel container θ1=20°C

Final temperature of liquid and steel containers θ2= 100°C

Mass of container M1 =15kg

Mass of liquid M2= 17kg

Specific heat capacity of steel container cst450 J/kg.C

Specific heat capacity of

Liquid cli = 4100 J/kg.C

We know that the heat applied is given as

Q= mcΔθ

Total heat applied = Ql + Qs

Where Ql= heat applied to liquid

Qs= heat applied to steel

Substituting our given data we have

Q= 15*450(100-20)+17*4100(100-20)

Q= 6750(80)+69700(80)

Q=540000+5576000

Q= 6116000

Q= 6116kJ

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5. A non-cold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa. Estimate the yield strength o
Papessa [141]

Answer:

97.17 MPa

Explanation:

Given:-

- The nominal strength of the grain, σ0  = 25 MPa

- The average grain size of the brass specimen, d* = 0.01 m

- The yield strength of the non-cold worked specimen, σy = 150 MPa

- Conditions of cold-working: T = 500°C , t = 1000 s

Find:-

Estimate the yield strength of this alloy after cold - working process

Solution:-

- The nominal strength of the grain is a function of yield strength of the material, grain yield factor ( Ky ) and the grain size.

- the following relation is used to determine the grain strength:

                             σ0  = σy  - ( Ky / √( d ) )

- We will use the above relation to determine the grain yield factor ( Ky ) for the alloy as follows. Note: here we will use the average value of grain size:

                            Ky = ( σy  - σy )*√( d* )

                            Ky = ( 150 - 25 ) * √0.01

                            Ky = 12.5 MPa - √mm

- Now we will use the cold working conditions of T = 500 C and time of the process is t = 1000 s. We will look up the elongated size of the grain after the cold-working process in lieu with its yield factor ( Ky ). Use figure 7.25.

- The cold-worked grain size with the given conditions can be read off from the figure 7.25. The new size comes out to be d = 0.03 mm.

- We will again use the nominal grain strength relation expressed initially. And compute for the new yield strength of the cold-worked alloy.

                            σ0  = σy  - ( Ky / √( d ) )

                            σy = σ0 + ( Ky / √( d ) )

                            σy = 25MPa + ( 12.5 / √( 0.03 mm ) )

                            σy = 97.17 MPa

- We see that the yield strength of the alloy decreases after cold-working process. This happens because the cold working process leaves with inter-granular strain ( dislocation of planes ) in the material structure which results from the increase in grain size.

6 0
3 years ago
Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured
aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]

If \bar{x} is 190, we can find the value of z\sigma_{\bar{x}}:

\bar{x} - z\sigma_{\bar{x}}  = 188.29

190 - z\sigma_{\bar{x}}  = 188.29

z\sigma_{\bar{x}}  = 1.71

Now we need to find the value of \sigma_{\bar{x}}:

\sigma_{\bar{x}} = s / \sqrt{n}

\sigma_{\bar{x}} = 5/ \sqrt{25}

\sigma_{\bar{x}} = 1

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

CI = 1 - 2*0.0436 = 0.9128 = 91.28\%

The nearest CI in the options is 90%, so the correct option is B.

4 0
3 years ago
There is a dispute between the multiple parties storing financial transaction data on a blockchain over the validity of a transa
spayn [35]

Answer:

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Explanation:

hjklllgfddsssssyjjjkkkk

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5 0
2 years ago
Why is concrete on its own not a good material to use
Inga [223]

Answer:

It has poor tensile strength despite having high compressive strength

Explanation:

Concrete exhibits high compressive strength when used. However, it has very low compressive strength. This is the reason why concrete is normally combined with steel to make a composite building material called reinforced concrete. The steel reinforces concrete hence increasing the tensile strength in RC buildings. The end composite is durable and fireproof. Generally, the main reason why concrete is not use on its own is due to its poor tensile strength.

7 0
2 years ago
14. A large car fire presents the possibility of
dexar [7]

Answer:

Both of the above

Explanation:

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2 years ago
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