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2 years ago

8

True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'

s Law.

1 answer:

defon2 years ago

8 0

Answer:

true

Explanation:

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Acquisition of resources from an external source is called?

densk [106]

Answer:

subcontracting

Explanation:

I hope this is right

6 0

2 years ago

A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago

Define a separate subroutine for each of the following tasks respectively.

Valentin [98]

Answer:

I HAVE NO CLUEEEE

Explanation:

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5 0

2 years ago

Read 2 more answers

Machine movement can be divided into what two main categories?

pishuonlain [190]

Answer:

motion and power

Explanation:

8 0

3 years ago

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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

2 years ago