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Report the answer using the correct number of significant figures! 1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0 Heat offusion of ic 100.0 g|79.7 cal/- 1970

Report the answer using the correct number of significant figures! 1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0 Heat offusion of ic 100.0 g|79.7 cal/- 1970

Chemistry
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? Mass of ice = 100.0g 100,0 g 79.7 cal-1970 cal 2. How much energy is required to vaporize 234.5 g of water? 234.5g 540 cal - 30.6 cal 259 = Heat of fusion of ice: 79.70 1.224 1.2 cal/g 1266301 1.2663X105 1.2663×10 1.3 x 10³ 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 mass of substance: 25g Heat vaporization: 30.6 cal mav that substance? Mass of 234.59 Heat of vaporization & 540 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? -550 cal mass m)- 500.09 AT=-1.10% 500.0g I cal ) (1.1008) (4.1845) (110) -2301.2₁-2,3012×10³. 11.10-2.30X 10³5 500.g 5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to 26.00 °C? 1000.08 | Cal = 3.00x/0³cal) 3% 8° 3.00 x 10³ cal 4, 1845/1.2552×10²4 :79.7 cal/g Answer: 7970 cal 5=25.65 Q==ssocal Answer: -550. cal (or -2.30 x 10³ J) Içal 6. The specific heat of copper is (0.0924 cal/g°C), how much energy temperature of 10.0 g of copper by 100.0 °C? 10.09 (0.0924 cal до с 25165=786gx sx35°C-20°c) +2=26.00°C 1.26 X10 450 x 10³ cal (or 1.26 x 10¹J) m= 1000.09 Q=S.M.AT S=1 cal/goc Teinal - Tinitial 26.00-23.00=3% T₁ = al) (100.0%) = ( 92.4 = is required to raise the mass of copper: 100g Specific heat copper: 0.0924cal/g °C AT 10 -100.0°C 7869 x 15% 0.00217² 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)? = mxsx Delta + mass(m) = 786 gram T₁ = 2010%² T₂ = 35.0°C Answer: 92.4 cal Answer: 2.17 x 10-³J/g °C -0.002171332 TFinal-Tinitial 35.0%-2010°C 15.0% 2617X10 J/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? Mass of ice = 100.0g 100,0 g 79.7 cal-1970 cal 2. How much energy is required to vaporize 234.5 g of water? 234.5g 540 cal - 30.6 cal 259 = Heat of fusion of ice: 79.70 1.224 1.2 cal/g 1266301 1.2663X105 1.2663×10 1.3 x 10³ 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 mass of substance: 25g Heat vaporization: 30.6 cal mav that substance? Mass of 234.59 Heat of vaporization & 540 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? -550 cal mass m)- 500.09 AT=-1.10% 500.0g I cal ) (1.1008) (4.1845) (110) -2301.2₁-2,3012×10³. 11.10-2.30X 10³5 500.g 5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to 26.00 °C? 1000.08 | Cal = 3.00x/0³cal) 3% 8° 3.00 x 10³ cal 4, 1845/1.2552×10²4 :79.7 cal/g Answer: 7970 cal 5=25.65 Q==ssocal Answer: -550. cal (or -2.30 x 10³ J) Içal 6. The specific heat of copper is (0.0924 cal/g°C), how much energy temperature of 10.0 g of copper by 100.0 °C? 10.09 (0.0924 cal до с 25165=786gx sx35°C-20°c) +2=26.00°C 1.26 X10 450 x 10³ cal (or 1.26 x 10¹J) m= 1000.09 Q=S.M.AT S=1 cal/goc Teinal - Tinitial 26.00-23.00=3% T₁ = al) (100.0%) = ( 92.4 = is required to raise the mass of copper: 100g Specific heat copper: 0.0924cal/g °C AT 10 -100.0°C 7869 x 15% 0.00217² 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)? = mxsx Delta + mass(m) = 786 gram T₁ = 2010%² T₂ = 35.0°C Answer: 92.4 cal Answer: 2.17 x 10-³J/g °C -0.002171332 TFinal-Tinitial 35.0%-2010°C 15.0% 2617X10 J/g °C
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