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7 months ago

Part A

Physics

1 answer:

Basile [38]7 months ago

6 0

The relationships to determine the number of calories to change 0.50 kg of 0°C ice to 0°C ice water is 1,080,000 cal.

<h3>How does heating ice that is at C affect it?</h3>

Ice melts and becomes liquid water at 0 degrees Celsius. Once all of the ice has been entirely transformed into liquid water, the temperature of the remaining ice begins to increase once more (in °C), continuing to rise until it reaches 100 °C, where it then stabilizes.

The water turns into steam when it reaches a temperature of 100 °C (D).

Water has a fusion latent heat of fusion of 80 cal/g.

Water has a 1 cal/g-C specific heat.

Water has a 540 cal/g latent heat of vaporization.

In light of this, the total amount of heat needed is 1500 g [(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g [(720 cal/g)] = 1,080,000 cal.

To learn more about Vaporization refer to:

brainly.com/question/26306578

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2 years ago

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pentagon [3]

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a) 1.55 cm

b) 4.25 cm

Explanation:

Given:

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$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

on substituting the values, we have

$\frac{1}{-2.60}=\frac{1}{5.80-v}+\frac{1}{v}$

on rearranging, we get

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on solving the quadratic equation, we get

v = 1.55 cm

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