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Mariulka [41]
2 years ago
15

How much heat is given off when 210.0g of water at 0 degrees freezes into ice?

Physics
1 answer:
Sergio [31]2 years ago
5 0

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

<h3>What is freezing?</h3>

It is a physical change in which liquids give off heat to form solids.

We have 210.0 g of water at 0°C. We can calculate the amount of heat given off when it freezes into ice using the following expression.

Q = ΔH°fus × m

Q = 0.334 kJ/g × 210.0 g = 70.1 kJ

where,

  • Q is the heat released.
  • ΔH°fus is the latent heat of fusion.
  • m is the mass.

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

Learn more about freezing here: brainly.com/question/40140

#SPJ1

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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine
Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = \frac{4}{3}\pi(\frac{d}{2})^3

or

Volume of the balloon = \frac{4}{3}\pi(\frac{0.296}{2})^3

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314  kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = \frac{3}{2}\times K_BT

where,

Boltzmann constant, K_B=1.3807\times10^{-23}J/K

Average kinetic energy = \frac{3}{2}\times1.3807\times10^{-23}\times292

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = \frac{3RT}{m}

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = \frac{3\times8.314\times292}{0.004}

or

rms speed = 1349.35 m/s

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Answer:

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