Calculate the amount of heat released when 150 grams of steamat 105 degree C is
ID: 691120 • Letter: C
Question
Calculate the amount of heat released when 150 grams of steamat 105 degree C is converted to ice at -12 degreeC. Use the information in the table below in your calculations asnecessary. Constant Value Specific heat ofwater 4.184J/g degree C Specific heat ofsteam 2.03 J/g degree C Specific heat ofice 2.09J/g degree C Heat of fusion ofice 333 J/g Heat of Vaporization ofwater 2260J/g Calculate the amount of heat released when 150 grams of steamat 105 degree C is converted to ice at -12 degreeC. Use the information in the table below in your calculations asnecessary. Constant Value Specific heat ofwater 4.184J/g degree C Specific heat ofsteam 2.03 J/g degree C Specific heat ofice 2.09J/g degree C Heat of fusion ofice 333 J/g Heat of Vaporization ofwater 2260J/gExplanation / Answer
Amount of heat released , Q = mcdt + mL + mc'dt + mL' +mc"dt" = m(cdt + L + c'dt' + L' + c"dt" ) Where m = mass of water = 150 g c =Specific heat of steam = 2.03 J/g degree C c' = Specificheat of water= 4.184 J/g degree C c" = Specificheat of ice= 2.09 J/g degree C L = Heat ofVaporization of water = 2260 J/g L'= Heat offusion of ice = 333 J/g dt = 105-100 =5oC dt' = 100 -0 =100oC dt" = 0-(-12)=12 oC Plug the values we get Q = m(cdt + L + c'dt '+ L'+ c"dt" ) = 456994.5 J = 456.9945 KJRelated Questions
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