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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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10. True or False? A disruptive technology<br> radically changes the way people live and<br> work.
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If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

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moment of inertia I = 144 inc^4

y = 3.96 in

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putting all value to get bending stress

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3 0
4 years ago
Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest gen
Anvisha [2.4K]

Answer:

a) Sample size = 1691

b) 95% Confidence Interval = (0.3696, 0.4304)

Explanation:

(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?

The margin of error is given by

MoE = z \cdot \frac{\sqrt{p(1-p)} }{\sqrt{n} }  \\\\

Where z is the corresponding z-score for 90% confidence level

z = 1.645 (from z-table)

for p = 0.50 and 2% margin of error, the required sample size would be

n = \frac{1.645^{2} \cdot 0.50(1-0.50)}{0.02^{2}}  \\\\n = \frac{0.6765}{0.0004}  \\\\n = 1691\\

(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find  a 95% confidence interval for p.

The sample proportion is

p = 400/1000

p = 0.40

z = 1.96 (from z-table)

n = 1000

The confidence interval is given by

CI = p \pm z \cdot \sqrt{\frac{p(1-p)}{n} } \\\\CI = 0.40 \pm 1.96 \cdot \sqrt{\frac{0.40(1-0.40)}{1000} } \\\\CI = 0.40 \pm 1.96 \cdot 0.01549 \\\\CI = 0.40 \pm 0.0304 \\\\CI = 0.40 - 0.0304 \: and \: 0.40 + 0.0304\\\\CI = (0.3696 ,\:  0.4304)

Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.  

3 0
4 years ago
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