Asked by Polina
26.5 kJ of heat is added to 5.44 g of ice at 0 C. What phase(s) are present and what is the temperature of the system after all the heat is absorbed?
Answers
Answered by
DrBob222
26,500 J heat initially.
To melt all of the ice we use up mass ice x heat fusion ice = 5.44g x 334 J/g = 1817 J which leaves 26,500- 1817 = 24683 J for heating the water at zero to some higher T.
To raise T of zero H2O to 100 C requires
mass x specific heat H2O x delta T =
5.44 x 4.184 x 100 = 2,276 J. Now we are left with 24683-2276 = 22,407 to boil the water and turn it into steam.
To turn all of it into steam requires mass H2O x heat vap = 5.44 x 2260 J/g = 12,294 J. Now we have left 22,407-12,294 = 10,113 left
Then q = mass x specific heat x delta T
10,113 J = 5.44 x 2 J/g*C x (Tf-100) = and solve for Tfinal.
Check all of these numbers.
To melt all of the ice we use up mass ice x heat fusion ice = 5.44g x 334 J/g = 1817 J which leaves 26,500- 1817 = 24683 J for heating the water at zero to some higher T.
To raise T of zero H2O to 100 C requires
mass x specific heat H2O x delta T =
5.44 x 4.184 x 100 = 2,276 J. Now we are left with 24683-2276 = 22,407 to boil the water and turn it into steam.
To turn all of it into steam requires mass H2O x heat vap = 5.44 x 2260 J/g = 12,294 J. Now we have left 22,407-12,294 = 10,113 left
Then q = mass x specific heat x delta T
10,113 J = 5.44 x 2 J/g*C x (Tf-100) = and solve for Tfinal.
Check all of these numbers.
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