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vfiekz [6]
3 years ago
5

Liquid sodium can be used as a heat transfer fluid. Its vapor pressure is 40.0 torr at 633°C and 400.0 torr at 823°C. Calculate

its heat of vaporization.
Physics
1 answer:
kaheart [24]3 years ago
4 0

Answer:

H vaporization = 100.0788 kJ/mol

Explanation:

Use clausius clapyron's adaptation for the calculation of Hvap as:

ln\frac {P_2}{P_1}=\frac {H_{vap}}{R}(\frac {1}{T_1}-\frac {1}{T_2})

Where,

P₂ and P₁ are the pressure at Temperature T₂ and T₁ respectively.

R is the gas constant.

T₂ = 823°C

T₁ = 633°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So, the temperature,

T₂ = (823 + 273.15) K = 1096.15 K

T₁ = (633 + 273.15) K = 906.15 K

P₂ = 400.0 torr , P₁ = 40.0 torr

R = 8.314 J/K.mol

Applying in the formula to calculate heat of vaporization as:

ln \frac {400}{40}=\frac {H_{vap}}{R} (\frac {1}{906.15}-\frac {1}{1096.15})

Solving for heat of vaporization, we get:

H vaporization = 100078.823 J/mol

Also, 1 J = 10⁻³ kJ

So,

<u>H vaporization = 100.0788 kJ/mol</u>

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A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g
Sliva [168]
To solve for force, you need to get the product of mass and acceleration. 
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration. 
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

a =  \frac{vf-vi}{t}

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s. 

So we can put it into our formula now:

a = \frac{0m/s-50m/s}{20ms}

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s. 

20ms x \frac{1s}{1000ms} = \frac{20s}{1000} = 0.02s

So your new time is 0.02s. Now we put this time into the formula:


a = \frac{0m/s-50m/s}{0.02s}
a =  \frac{-50m/s}{0.02s}  = -2,500 m/ s^{2}

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop. 

So now we have our acceleration. Now using this, we can get our force. 

F= ma

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is kg.m/ s^{2} but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first. 

150g =  \frac{1kg}{1,000g}  =  \frac{150kg}{1,000}  = 0.150kg

Our new mass is 0.150kg. 

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 m/ s^{2}

Now that all our units match, we can put that into our formula:

F= ma
F= (0.150kg)(-2,500m/s^{2})
F = -375kg.m/ s^{2}  or -375N

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign. 



4 0
3 years ago
Which layer of the earths atmosphere contains the ozone layer?
Burka [1]
The answer to that will be the Troposphere.
8 0
3 years ago
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a ball of mass 0.5 kg moving at 10 m/s collides with another ball of equal mass at rest. if the two balls move off together afte
nexus9112 [7]

Answer:

5 m/s

Explanation:

Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.

Initial momentum = Final momentum

0.5 × 10  = 1 × V

V = 5 m/s

7 0
3 years ago
A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-
musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0
3 years ago
A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta
qwelly [4]

Answer:0.0133 \Omega

Explanation:

Given

Voltage=12 V

Current(I)=150 A

V_{terminal}=10 V

r_{internal}=\frac{\Delta V}{I}

r_{internal}=\frac{12-10}{150}

r_{internal}=\frac{2}{150}=0.0133 \Omega

4 0
3 years ago
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