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RSB [31]
3 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]3 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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miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

4 0
4 years ago
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
4 years ago
A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
Mazyrski [523]

Answer:

P > 142.5 N  (→)

the motion sliding

Explanation:

Given

W = 959 N

μs = 0.3

If we apply

∑ Fy = 0 (+↑)

Ay + By = W

If  Ay = By

2*By = W

By = W / 2

By = 950 N / 2

By = 475 N (↑)

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F = μs*N = μs*By

F = 0.3*475 N

F = 142.5 N (←)

we can apply

P - F  > 0

P  > 142.5 N (→)

the motion sliding

6 0
3 years ago
Water enters a boiler at a temperature of 110°F. The boiler is to produce 2000 lb/hr of steam at a pressure of 130 psia. How man
strojnjashka [21]
Yes it does the answer is no
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2 years ago
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Height is a quantitative variable because it is related to the measurement and in measurement, when we measure something we deal with number (numerical data)

Numerical data is a type of quantitative data that is why we say “height is a quantitative variable”  

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<u>Answer: </u>The brand name and model  

8 0
3 years ago
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