Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Can you help me with the number 5 questions? Also can you please show the step-by-step including the formula?
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 Heat of fusion of ice: 79. 100.0g|79.7.cal-1970 cal 2. How much energy is required to vaporize 234.5 g of water? 234.5g 540 cal- - 1.224 1.2 cal/g 12 6630 1.2663X105 1.3x105 Answer: 1.3 x 105 cal cal 3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of that substance? mass of substance: 25g energy require vaporization: 30.6 Cal may 30.6 cal 259 26.00 °C? 100c,05 (6બ (22)=xtel 3.00x 10³cal 4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C? mass m) = 500.09 AT=-1.10°C 500.0g (ca.1000) = [ 500.g (4.1845 -1.10 23.01.2-2,3012×10³. -550 cal 5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to 0.0924 cal 3-21) (100.024) до с m= 1000.09 Q=SOMIAT S=1 cal/goc Teinal-Tinitid T₁ = 23.00% T2=26.00°C 3.00 x 10³ cal (or 1.26 x 10¹J) is required to raise the mass of copper: 100g Specific heat copper: 0.0924cal/g °C AT=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)? = mxsx Delta + mass(m) = 786 gram T₁ = 20.0% T₂=35:0°C -0.002171332 TFinal-Tinitial 0.00217 4184511.2552x10 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 Mass of water 234.59 Heat of vaporization: 540 cal/g Heat of vaporization substance: 1.2 cal/g Answer: 1.2 cal/g 25.65=786 gx sx35°C-20°c) ---5=25.6J 7869x :79.7 cal/g Answer: 7970 cal 15°C Q==ssocal Answer: -550. cal (or -2.30 x 10³ J) 92.4call (2617X10³J/g °C 26.00-23.00=3% Answer: 2.17 x 10-³J/g °C 35.0%-2010°C = 15.0°C
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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 Heat of fusion of ice: 79. 100.0g|79.7.cal-1970 cal 2. How much energy is required to vaporize 234.5 g of water? 234.5g 540 cal- - 1.224 1.2 cal/g 12 6630 1.2663X105 1.3x105 Answer: 1.3 x 105 cal cal 3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of that substance? mass of substance: 25g energy require vaporization: 30.6 Cal may 30.6 cal 259 26.00 °C? 100c,05 (6બ (22)=xtel 3.00x 10³cal 4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C? mass m) = 500.09 AT=-1.10°C 500.0g (ca.1000) = [ 500.g (4.1845 -1.10 23.01.2-2,3012×10³. -550 cal 5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to 0.0924 cal 3-21) (100.024) до с m= 1000.09 Q=SOMIAT S=1 cal/goc Teinal-Tinitid T₁ = 23.00% T2=26.00°C 3.00 x 10³ cal (or 1.26 x 10¹J) is required to raise the mass of copper: 100g Specific heat copper: 0.0924cal/g °C AT=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)? = mxsx Delta + mass(m) = 786 gram T₁ = 20.0% T₂=35:0°C -0.002171332 TFinal-Tinitial 0.00217 4184511.2552x10 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 Mass of water 234.59 Heat of vaporization: 540 cal/g Heat of vaporization substance: 1.2 cal/g Answer: 1.2 cal/g 25.65=786 gx sx35°C-20°c) ---5=25.6J 7869x :79.7 cal/g Answer: 7970 cal 15°C Q==ssocal Answer: -550. cal (or -2.30 x 10³ J) 92.4call (2617X10³J/g °C 26.00-23.00=3% Answer: 2.17 x 10-³J/g °C 35.0%-2010°C = 15.0°C
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