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Anastaziya [24]

3 years ago

12

A rigid, insulated tank that is initially evacuated is connected though a valve to a supply line that carries steam at 1 MPa and

300∘C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed.
Determine the final temperature of the steam in the tank, in ∘C.

1 answer:

meriva3 years ago

6 0

Answer:

Final Temperature of the steam tank = 456.4°C

Explanation:

Assuming it to be a uniform flow process, kinetic and potential energy to be zero, and work done and heat input to be zero also. We can conclude that,

Enthalpy of the steam in pipe = Internal Energy of the steam in tank

Using the Property tables and Charts - Steam tables,

At Pressure= 1 MPa and Temperature= 300°C,

Enthalpy = 3051.2 kJ/kg

At Pressure= 1 MPa and Internal Energy= 3051.2 kJ/kg,

Temperature = 456.4°C.

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The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.6 V and b = -4.90 V/m. (a) Determine the pote

cestrela7 [59]

Answer:

a) 10.6V

b) E = 4.9V/m, +x direction

c) E = 4.9V/m, +x direction

Explanation:

You have the following function:

$V=a+bx$ (1)

for the potential in a region between x=0 and x=6.00 m

a = 10.6 V

b = -4.90V/m

$V=10.6V-4.90\frac{V}{m}x$

a) for x=0 you obtain for V:

$V=10.6V-4.90\frac{V}{m}(0m)=10.6V$

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4 years ago

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

B)

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

C)

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

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Answer:

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