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2 years ago

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

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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Answer:

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Explanation:

We know that the centripetal acceleration is

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where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

<h3>b. </h3>

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

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