A)
Ti = 17.0 oC
Tf = 0.0 oC
Cl = 4.184 J/g.oC
Heat released to convert liquid from 17.0 oC to 0.0 oC
Q1 = m*Cl*(Ti-Tf)
= 1.75 g * 4.184 J/g.oC *(17-0) oC
= 124.474 J
Hf = 334 J/g
Heat released to convert liquid to solid at 0.0 oC
Q2 = m*Hf
= 1.75g *334 J/g
= 584.5 J
Total heat released = Q1 + Q2
= 124.474 J + 584.5 J
= 708.974 J
Answer: 709 J
B)
Ti = 0.0 oC
Tf = 70.0 oC
here
Hf = 334 J/g
Heat required to convert solid to liquid at 0.0 oC
Q1 = m*Hf
= 40.0g *334 J/g
= 13360 J
Cl = 4.184 J/g.oC
Heat required to convert liquid from 0.0 oC to 70.0 oC
Q2 = m*Cl*(Tf-Ti)
= 40 g * 4.184 J/g.oC *(70-0) oC
= 11715.2 J
Total heat required = Q1 + Q2
= 13360 J + 11715.2 J
= 25075.2 J
Answer: 25100 J
C)
Ti = 100.0 oC
Tf = 0.0 oC
Hv = 2260 J/g
Heat released to convert gas to liquid at 100.0 oC
Q1 = m*Hv
= 11.0g *2260 J/g
= 24860 J
Cl = 4.184 J/g.oC
Heat released to convert liquid from 100.0 oC to 0.0 oC
Q2 = m*Cl*(Ti-Tf)
= 11 g * 4.184 J/g.oC *(100-0) oC
= 4602.4 J
Total heat released = Q1 + Q2
= 24860 J + 4602.4 J
= 29462 J
= 29.5 KJ
Answer: 29.5 KJ
D)
Ti = 0.0 oC
Tf = 100.0 oC
here
Hf = 334 J/g
Heat required to convert solid to liquid at 0.0 oC
Q1 = m*Hf
= 23.5g *334 J/g
= 7849 J
Cl = 4.184 J/g.oC
Heat required to convert liquid from 0.0 oC to 100.0 oC
Q2 = m*Cl*(Tf-Ti)
= 23.5 g * 4.184 J/g.oC *(100-0) oC
= 9832.4 J
Hv = 2260 J/g
Heat required to convert liquid to gas at 100.0 oC
Q3 = m*Hv
= 23.5g *2260 J/g
= 53110 J
Total heat required = Q1 + Q2 + Q3
= 7849 J + 9832.4 J + 53110 J
= 70791.4 J
= 70.8 KJ
Answer: 70.8 KJ
Using the heat of fusion for water 334 J/g the heat of vaporization for water 2260...
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<Chapter 3 Homework Problem 3.102 Part A .1 Review 1 Constants l Periodic Table A 111 g sample of steam at 100 °C is emitted from a volcano. It condenses, cools, and falls as snow at 0°C How many kilojoules of heat were released? Express your answer to three significant figures and include the appropriate units Units Submit X Incorrect; Try Again; 5 attempts remaining < Return to AssignmentProvide Feedback
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