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Amiraneli [1.4K]
2 years ago
6

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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The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

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now according to principal of conservation of energy we observe,

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μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

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Power is your answer :)

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Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

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In order to find the electric flux we first have to find out the area of triangle.

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