At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

Answers

Answer 1

Answer:

The molar latent enthalpy of boiling of iron at 3330 K is ΔH = 342 $*$ 10^3 J.

Explanation:

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

d ln p = (ΔH / RT^2) dt

(1/p) dp = (ΔH / RT^2) dt

dp / dt = p (ΔH / RT^2) = 3.72 $*$ 10^-3

(p) (ΔH) / (8.31) (3330)^2 = 3.72 $*$ 10^-3

ΔH = 342 $*$ 10^3 J.

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A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min

Answers

Answer:

6.4 rpm

Explanation:

$I_(m)$ = moment of inertia of merry-go-round = 275 kgm²

m = mass of the child = 23 kg

R = radius of the merry-go-round = 2.20 m

$I_(c)$ = moment of inertia of child after jumping on merry-go-round = mR² = (23) (2.20)² = 111.32 kgm²

Total moment of inertia after child jumps is given as

$I_(f)$ = $I_(m)$ + $I_(c)$ = 275 + 111.32 = 386.32 kgm²

Total moment of inertia before child jumps is given as

$I_(i)$ = $I_(m)$ = 275 kgm²

$w_(i)$ = initial angular speed = 9 rpm

$w_(f)$ = final angular speed

using conservation of angular momentum

$I_(i)$ $w_(i)$ = $I_(f)$ $w_(f)$

(275) (9) = (386.32) $w_(f)$

$w_(f)$ = 6.4 rpm

Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?

Answers

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before shooting over Niagara Falls. After moving over the edge, the water drops 53 mm to the water below.a. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom?
b. Express your answer to two significant figures and include the appropriate units.
c. How far does the water move horizontally during this time?
d. Express your answer to two significant figures and include the appropriate units.

Answers

Answer :

a.3.29 m/s

b.3.3 m/s

c.14.8 m

d.15 m

Explanation:

We are given that

Initial Horizontal speed= $v_x=4.5 m/s$

Vertical component of initial speed= $v_y=0$

Vertical distance= $y=-53 m$

a. $s=ut+(1)/(2)gt^2$

Using the formula and g is negative therefore $g=-9.8m/s^2$

$-53=0-(1)/(2)(9.8)t^2$

$53=4.9t^2$

$t^2=(53)/(4.9) s$

$t=\sqrt{(53)/(4.9)}=3.29m/s$

b.Hundredth place is greater than 5 therefore, 1 will be added to tenth place and other digits on left side of tenth place remains same and digit on right side of tenth place replace by 0.

$t=3.3 m/s$

c.Horizontal acceleration= $a_x=0$

$x=v_xt=4.5* 3.29=14.8 m$

d.Tenth place 8 is greater than 5 therefore, 1 will be added to unit place and other digits on left side of unit place remains same and digit on right side of unit place replace by 0.

Horizontal distance=15 m

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?

Answers

A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $(2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) ) = (106x)/(x+53)$

$26.5 = (106x)/(x+53) \n \n 79.5 x = 1404.5\n \n x = 17.67 miles/hour$

Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?