Asked by Mindy
Calculate the amount of heat, in kJ, needed to convert 30.0 g of ice at -5.2 oC to liquid water at 58 oC.
Specific heat of ice = 2.10 J/(g oC); heat of fusion of ice = 334 J/g; Specific heat of water = 4.18 J/(g oC)
Specific heat of ice = 2.10 J/(g oC); heat of fusion of ice = 334 J/g; Specific heat of water = 4.18 J/(g oC)
Answers
Answered by
DrBob222
q1 = heat needed to warm ice at -5.2 C to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tf is zero and Ti is -5.2.
q2 = heat needed to change phase from solid ice at zero to liquid water at zero.
q2 = mass ice x heat fusion
q3 = heat needed to warm liquid H2O at zero C to 58 c.
q3 = mass liquid water x specific heat liq water x (Tf - Ti)
Total q = q1 + q2 + q3.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tf is zero and Ti is -5.2.
q2 = heat needed to change phase from solid ice at zero to liquid water at zero.
q2 = mass ice x heat fusion
q3 = heat needed to warm liquid H2O at zero C to 58 c.
q3 = mass liquid water x specific heat liq water x (Tf - Ti)
Total q = q1 + q2 + q3.
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