My teacher did not explain to me how to do this kind ofproblem.. help is much ap

Question

My teacher did not explain to me how to do this kind ofproblem.. help is much appreciated. "You have a 46 gram sample of H20 at a temperatureof -58 degrees C. How many joules of energy are necessary to raisethe temperature of the sample to 114 degrees C?" The specific heat capacity of ice is 2.06 J/g deg.C The specific heat capacity of water is 4.18 J/g deg.C The specific heat capacity of steam is 2.02 J/gdeg.C The heat of fusion of ice is 334 J/g The heat of vaporization is 2260 J/g My teacher did not explain to me how to do this kind ofproblem.. help is much appreciated. "You have a 46 gram sample of H20 at a temperatureof -58 degrees C. How many joules of energy are necessary to raisethe temperature of the sample to 114 degrees C?" The specific heat capacity of ice is 2.06 J/g deg.C The specific heat capacity of water is 4.18 J/g deg.C The specific heat capacity of steam is 2.02 J/gdeg.C The heat of fusion of ice is 334 J/g The heat of vaporization is 2260 J/g

Explanation / Answer

Given mass , m = 46 g The specific heat capacity of ice is c = 2.06 J/g deg.C The specific heat capacity of water is C = 4.18 J/gdeg.C The specific heat capacity of steam is C' =2.02 J/gdeg.C The heat of fusion of ice is L = 334 J/g The heat of vaporization is L' = 2260 J/g The required heat , Q = mcdt + mL + mCdt ' + mL ' + mC ' dt"
                                = m( cdt + L + Cdt ' + L ' + C ' dt " ) where m = 46 g dt = 0-(-58 ) = 58 oC dt'= 100 - 0 =100 oC dt" = 114 - 100 = 14 oC Plug the values we get Q =  m(cdt + L + Cdt ' + L ' + C ' dt " )                                       = 46* ( 119.48 + 334 + 418 + 2260 + 28.28 )                                        =145348.96 J

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