The Expert TA | Human-X e Chegg Study I Guided Solut + LTX https: /usx50wvtheexp

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The Expert TA | Human-X e Chegg Study I Guided Solut + LTX https: /usx50wvtheexpertta.com Common TakeTutorialssignment.aspx Hae Student: klantzS94cardinal.wju.edu My Account Log Out I Halp Ch14 Calorimetry Begin Date: 119:201812:01:00 AM -Due Dat23 2018 11:59:00 PM Fnd Date: 1262018 11:59:00 PM (20%) Problem 1: A block of aluminum ata temperature of 1,-30.5 degrees C has a mass of m-11 kg. It is brought into coulact with anoher block of alumiuum with mass ol m 0.5ms, al a emperalure of I 14 degrees C. This alumiuum has a heat capacity ofc 899 J/kg K Assignment Status Click here for detailed view 33% Part (a) Input an expression for the final temperature of the blocks. Grade Summary 0% 100 Problem Status 7 8 9 4 5 6 Snbmisslons Attents remanin (2% per attempt) detatled vie Completed Completed Completed Partial Ti Subeint I Tve Hints: 3 fnrn 0% deduction Hints Foodhack: dhtion per foodback The system ia isolated, ao what muat the total change in beat of the ystem ba The heat lost hy onc block must be the hcat pained by the other -Set up this relation and solve for the final 33% Part (b) What is the lempera lure in degrees C? @ ee 33% Part (c) what was the change in heat of block 1, in J? 9:04 PM Type here to search 1/21/2018

Explanation / Answer

Heat lost by 1st block = heat gained by 2nd block

m1*c*(T1 - Ts ) = m2*c*(Ts - T2)

(30.5-T)= 0.5*(T-14)

30.5-T= 0.5T - 7

37.5= 1.5*T

T= 25 degree Celsius.

This is the new equilibrium temperature.

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