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(2) New messages from M Blackboard Learn xbboard2010.nwosu.edu/b xE-Mail Access Northwest xta The Expert TA | Human-lik Northwestern Oklahoma x bboard2010.nwosu.edu/t + E-Mail Access l Northwes. × x × uman-lik × -) Dwww.saplinglearning.com/ib.scms/mod/ibis/view.php?id=1960309 Jump to... Logout ? Help sapling learning Sapling Learning My Assignment Northwestern Oklahoma State University-CHEM 1115-Fall15-WICKHAM Activities and Due Dates HW 6a Resources 10/7/2015 11:55 PM o/100 Gradebook O Assignment Information Available From: 9/29/2015 06:00 PM Due Date Points Possible: 100 Grade Category: Graded Description: Policies: Attempts Score Print PrintCalculatorPeriodic Table Calculator Question 1 of 5 10/7/2015 11:55 PM Mapdo sapling leaming If a system has 4.00 x 102 kcal of work done to it, and releases 5.00 x 102 kJ of heat into its surroundings what is the change in internal energy of the system? You can check your answers. Number You can view solutions when you complete or give up on any question You can keep trying to answer each question until you get it right or give up. You lose 5% of the points available to each answer in your question for each incorrect attempt at that kJ OiHelp With This Topic OlWeb Help & Videos Technical Support and Bug Reports . Previous Give Up & View Solution 0 Check Answer Next Exit Copyright © 2011-2015 Sapling Learning, Inc.-158 about uscareers partners privacy polilcy rms of usecoct us help 1:35 AM Search the web and Windows O e C x d 10/7/2015Explanation / Answer
Change in internal energy deltaE is given by formula,
deltaE = q + w
where,
q = -5 x 10^2 kJ [heat released to the surrounding is -ve]
w = 4 x 10^2 kcal = 1673.6 kJ [work done on the system is +ve]
So we get,
deltaE = -5 x 10^2 + 1673.6 = 1173.6 kJ
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