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3 years ago

Question 8(Multiple Choice Worth 2 points)

Engineering

1 answer:

8090 [49]3 years ago

8 0

Answer:

4 number answer is correct.

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A 4 cm diameter sphere of copper is initially at a temperature of 95 °C. It is placed in a very large water bath at time t- 0. T

avanturin [10]

Answer:112.376 s

Explanation:

Given

$T_i=95^{\circ}C$

$T_f=35^{\circ}C$

$T_\infty \left ( ambient\right )=25^{\circ}C$

$h=400 watts/\left ( m^{2}^{\circ}C\right )$

$c=0.385 J/\left ( m^2^{\circ}C\right )$

$\rho =9 gm/cm^{3}$

Using Newton's law of cooling

$\frac{T_i-T_{\infty}}{T-T_{\infty}}$ = $e^{\frac{ht}{\rho L_{c}c}}$

$\frac{95-25}{35-25}$ = $e^{\frac{400\times 3\times 10^{-4}\times t}{9\times 2\times 0.385}}$

7= $e^{1.7316\times 10^{-2}\times t}$

Taking log both side

t=112.376sec

4 0

3 years ago

8. What is the density of an object with a mass of 290.5 g and volume of 83 cm 3?

Leni [432]

Answer:

The density would be 218.5

6 0

3 years ago

Hello, how are you?

Kisachek [45]

Answer:

Hello, I'm good. Thank you for asking

8 0

2 years ago

Read 2 more answers

In urban area you can except

liberstina [14]

Answer:

eoidnfoejsdncodsnc

Explanation:

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5 0

3 years ago

Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from

Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet ( $h_{1}=40.09$ Btu/lb)

liquid specific enthalpy at the exit ( $h_{2}=40.94$ Btu/lb)

initial elevation ( $z_1=0ft$ )

final elevation ( $z_2=100ft$ )

acceleration due to gravity (g) = 32.174 ft/s²

$W_{cv}$ = 3 Btu/s

The energy balance equation is given as:

$Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Since kinetic energy effects are negligible, the equation becomes:

$Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0$

Substituting values:

$Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\$

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0

3 years ago