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2 years ago

How many calories are released when 6 grams of 100°C steam turns to 0°C ice?

Physics

1 answer:

maks197457 [2]2 years ago

6 0

Answer: 4276.2 calories

Explanation:

Given

mass of steam is 6 gm at $100^{\circ}C$

Conversion of steam to ice involves

steam to water at $100^{\circ}C$

water at $100^{\circ}C$ to water at $0^{\circ}C$

water to the ice at $0^{\circ}C$

Calories released during the conversion of steam to water at $100^{\circ}C$

$E_1=mL_v\quad [L_v=\text{latent heat of vaporisation}]\\E_1=6\times 533=3198\ cal.$

Calories released during the conversion of water at $100^{\circ}C$ to water at $0^{\circ}C$

$E_2=mc(\Delta T)\quad [c=\text{specific heat of water,}1\ cal./gm.^{\circ}C]\\E_2=6\times 1\times 100=600\ cal.$

Calories released during the conversion of water to the ice at $0^{\circ}C$

$E_3=mL_f\quad [L_f=\text{latent heat of fusion}]\\E_3=6\times 79.7=478.2\ cal.$

The total energy released is

$E=E_1+E_2+E_3\\E=3198+600+478.2=4276.2\ cal.$

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A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about

Misha Larkins [42]

<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)

The capacitance of a series combination is

1 / (1/A + 1/B + 1/C + 1/D + .....) .

If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is

(product of the 2 individuals) / (sum of the individuals) .

In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is

(1000 x 1) / (1000 + 1)

= (1000) / (1001)

= 0.999 000 999 . . .

which is smaller than the smaller individual.

It'll always be that way. </span>

5 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

True or False: It would be easier to run on a planet with high gravity than one with less gravity.

LUCKY_DIMON [66]

True! It would be easier

3 0

3 years ago

Fiora starts riding her bike at 18 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the

lord [1]

Answer:

t = 2.5 hours

Explanation:

given,

speed of the bike for t time= 18 mi/h

final speed of the bike after t time = 12 mi/h

total distance, D = 69 miles

total time, T= 4.5 hour

time for which speed of the bike is 18 mi/h = ?

we know distance = speed x time

now,

18 x t + 12 (4.5 - t) = 69

6 t + 54 = 69

6 t = 15

t = 2.5 hours

The bike was at the speed of 18 mi/h for 2.5 hours.

4 0

3 years ago

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N

vovangra [49]

Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²

Answer: 0.665 m/s² (nearest thousandth)

7 0

3 years ago