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is owLv2 | Online teaching × () Chegg Study I Guided s x C cvg.cengagenow.com/ilrn/takeAssignment/takeCovalent Activity do?locator-assignment-take&takeAssignmentSessionLocator-assignment-take; : Apps : kober portal D 2004 Saturn L300 H: web chemucsbedu -Baltimore Classificat Acute Viral Hepatitis Hepatitis A Treatme Jaundice: MedlinePl Other bookmarks Time Remaining; 1:36:04 UNIT TEST Use the References to access important values if needed for this question. Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 1 pt 1 pt 1 pr 1 pt 1 pt 1 pt 1 pt 1 pr 1 pt 1 pt 1 pt 1 pt 1 pr 1 pt 1 pt 1 pt The following information is given for lead at 1 atm: AHap(1740.00°C) 858.2J g Aru' (328.00°C) = 23.00 Jg Tb 1740.00°C Tm 328.00°C Specific heat solid-0.1300 J/g C Specific heat liquid = 0.1380 Jg °C kJ of heat are needed to A 26.20 g sample of solid lead is initially at 302.00°C If the sample is heated at constant pressure (P-1 atm). raise the temperature of the sample to 571.00°C Submit Answer Try Another Version 1 item attempt remaining Question 19 Question 20 Question 21 Question 22 Question 23 Question 24 Question 25 Progress: 0/25 items Due Nov 2 at Previous Next 11:00 PM 8:48 PM 11/2/2017Explanation / Answer
Ti = 302.0
Tf = 571.0
here
Cs = 0.13 J/g.oC
Heat required to convert solid from 302.0 oC to 328.0 oC
Q1 = m*Cs*(Tf-Ti)
= 26.2 g * 0.13 J/g.oC *(328-302) oC
= 88.556 J
Lf = 858.2 J/g
Heat required to convert solid to liquid at 328.0 oC
Q2 = m*Lf
= 26.2g *858.2 J/g
= 22484.84 J
Cl = 0.138 J/g.oC
Heat required to convert liquid from 328.0 oC to 571.0 oC
Q3 = m*Cl*(Tf-Ti)
= 26.2 g * 0.138 J/g.oC *(571-328) oC
= 878.5908 J
Total heat required = Q1 + Q2 + Q3
= 88.556 J + 22484.84 J + 878.5908 J
= 23452 J
= 23.45 KJ
Answer: 23.45 KJ
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