a car has a mass of 200kg. It is on a hill 1000m high. How much gravitational potential energy does the car have?

200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.

balu736 [363]

Answer:

a. -0.01324 kJ/K, b. = 0.03233 kJ/K , c. = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)

3.54 = 0.1 . X x 203.29

x = 0.1741, which is the dryness fraction of fluid at the final state.

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= 20.329 kJ, which is much greater than 3.45 kJ, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.

6 0

3 years ago

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

3 years ago

two wires have the same length, but the second has twice the diameter of the first. find the resistance of the second wire if th

djverab [1.8K]

2ω is the resistance of the second wire if the resistance of the first is 4ω if two wires have the same length, but the second has twice the diameter of the first.

R= 4ω.

R = ρl/A

2d=r

R2=2ω

Resistance is the capacity of a conductor to obstruct the passage of an electric current through it. It is controlled by the interaction of the applied voltage and the electric current passing through it.

Conductors have very little resistance, whereas insulators have a significant amount of resistance. The resistance increases as the current flow decreases. Resistance is influenced by the properties and dimensions of the material (area of cross section)

To know more about resistance visit : brainly.com/question/14547003

#SPJ4

5 0

1 year ago

Hello please help i’ll give brainliest

balu736 [363]

D, I hope this is correct!

6 0

3 years ago

A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph

Nina [5.8K]

Think about how each of noise, heat, visible light, radiation and atmospheric shock waves travel. Which ones require air particles to travel?

A vacuum has no air particles within it, it is completely empty.

Therefore, any of the above that requiring particles to travel will not be able to cross it and the observer will not experience it.

8 0

3 years ago