Homework Help is Here – Start Your Trial Now!
Science
Chemistry
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)? 55,6J = 786g xs x(35°C -20°c) 5=25.65 2.171331637X10 (1869 x 15°C) (2.17 X10-²³5/9°C Tfinal-Tinitial 35.0°C-20.0°C = 15.0°0 Answer: 2.17 x 10³J/g °C

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)? 55,6J = 786g xs x(35°C -20°c) 5=25.65 2.171331637X10 (1869 x 15°C) (2.17 X10-²³5/9°C Tfinal-Tinitial 35.0°C-20.0°C = 15.0°0 Answer: 2.17 x 10³J/g °C

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 17CR: Are changes in state physical or chemical changes? Explain. What type of forces must be overcome to...
See similar textbooks
icon
Related questions
Question
100%

Can you help me number 7? Is this correct answer with show my work including the Significant Figures number?

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? 100.0 g 79.7 cal ·cal = 7970 cal 9 2. How much energy is required to vaporize 234.5 g of water? 1.2663 х 105 234.5g 540 cal g 1.3X105cal 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? 30.6 cal 259 1.224 1.2 cal/g 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? 500.09 1 cal) (-1.10°C) = [550 cal 500.g -2,3012×103 4.184J (-1.10) = √2.30×10³5 Answer: -550. cal (or -2.30 x 10³ J) I cal 5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to M=1000.09 Q=S•m. (OT) s=lcal/g °C T1=23.00% 26.00 °C? 1000,03 (1Cal) (3°C) = 5+00X10³ cal go c 1,2552x10 -1.26x1045 3,00x10³ cal 4.1845). 1461845-1-20 10.0924 cal 9°C Answer: 7970 cal 1) (100.0°C) = 12,4 cal 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? 10.09 T2= 26.00°C Answer: 3.00 x 10³ cal (or 1.26 x 10¹J) 25,65=786g xs x (35°C -20°c) 5=25.65 2.171331637X10 (7869 x 15°C) (2.17X/10-3³5/9°C Tfinal-Tinitial 26.00-23.00=3c -3 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)? Tfinal-Tinitial 35.0°C-20.0°= 15.00 Answer: 2.17 x 10-³J/g °C
Transcribed Image Text: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? 100.0 g 79.7 cal ·cal = 7970 cal 9 2. How much energy is required to vaporize 234.5 g of water? 1.2663 х 105 234.5g 540 cal g 1.3X105cal 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? 30.6 cal 259 1.224 1.2 cal/g 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? 500.09 1 cal) (-1.10°C) = [550 cal 500.g -2,3012×103 4.184J (-1.10) = √2.30×10³5 Answer: -550. cal (or -2.30 x 10³ J) I cal 5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to M=1000.09 Q=S•m. (OT) s=lcal/g °C T1=23.00% 26.00 °C? 1000,03 (1Cal) (3°C) = 5+00X10³ cal go c 1,2552x10 -1.26x1045 3,00x10³ cal 4.1845). 1461845-1-20 10.0924 cal 9°C Answer: 7970 cal 1) (100.0°C) = 12,4 cal 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? 10.09 T2= 26.00°C Answer: 3.00 x 10³ cal (or 1.26 x 10¹J) 25,65=786g xs x (35°C -20°c) 5=25.65 2.171331637X10 (7869 x 15°C) (2.17X/10-3³5/9°C Tfinal-Tinitial 26.00-23.00=3c -3 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)? Tfinal-Tinitial 35.0°C-20.0°= 15.00 Answer: 2.17 x 10-³J/g °C
Expert Solution
steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Thermodynamics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
Chemistry
ISBN:
9781337399425
Author:
Steven S. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry: The Molecular Science
Chemistry: The Molecular Science
Chemistry
ISBN:
9781285199047
Author:
John W. Moore, Conrad L. Stanitski
Publisher:
Cengage Learning
Principles of Modern Chemistry
Principles of Modern Chemistry
Chemistry
ISBN:
9781305079113
Author:
David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:
Cengage Learning
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781337399074
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781133949640
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning
Chemistry for Engineering Students
Chemistry for Engineering Students
Chemistry
ISBN:
9781337398909
Author:
Lawrence S. Brown, Tom Holme
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS