Can u help me with my homework Calculations for Temperature and Phase Change WorksheetThe heat of fusion of ice is 79.7 cal/g.The heat of vaporization of water is 540 cal/g.Report the answer using the correct number of significant figures!1. How much energy is required to melt 100.0 grams of ice?2. How much energy is required to vaporize 234.5 g of water?Answer: 7970 calAnswer: 1.3 x 105 cal3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization ofthat substance?Answer: 1.2 cal/g4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C?Answer: -550. cal (or -2.30 x 10³ J)5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to26.00 °C?Answer: 3.00 x 103 cal (or 1.26 x 104J)6. The specific heat of copper is (0.0924 cal/g°C), how much energy is required to raise thetemperature of 10.0 g of copper by 100.0 °C?Answer: 92.4 cal7. If 25.6 J of energy raised 786 g of a substance from 20.0°C to 35.0°C, what is the specific heat ofthe substance (S)?Answer: 2.17 x 10-³J/g °Celet

Answered: the correct number of significant…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Calculations for Temperature and Phase Change Worksheet
The heat of fusion of ice is 79.7 cal/g.
The heat of vaporization of water is 540 cal/g.
Report the answer using the correct number of significant figures!
1. How much energy is required to melt 100.0 grams of ice?
2. How much energy is required to vaporize 234.5 g of water?
Answer: 7970 cal
Answer: 1.3 x 105 cal
3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of
that substance?
Answer: 1.2 cal/g
4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C?
Answer: -550. cal (or -2.30 x 10³ J)
5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to
26.00 °C?
Answer: 3.00 x 103 cal (or 1.26 x 104J)
6. The specific heat of copper is (0.0924 cal/g°C), how much energy is required to raise the
temperature of 10.0 g of copper by 100.0 °C?
Answer: 92.4 cal
7. If 25.6 J of energy raised 786 g of a substance from 20.0°C to 35.0°C, what is the specific heat of
the substance (S)?
Answer: 2.17 x 10-³J/g °C
elet

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To solve this problem we will use the given data in the question .

Heat of fussion of ice = 79.7 cal/g

Heat of vaporization water = 540 cal/g

Note : According to Bartleby question answer guidelines we are supposed to solve first three questions given in the sheet as you don't have told to solve specific question , Thank you!

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