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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No sorry, wish I could help
Answer:
attribute is defined as a quality or characteristic of a person, place, or thing
Explanation:
82 ÷ 6.5 = 12.615384615384... repeating
round = 12.6
12.6 · 6.5 = 81.9
round = 82
She went an average of 12.6 km an hour
Hope this helps! ;)
To me, that sounds like the "Law of Conservation of Energy".
