Asked by fez
Calculate the enthalpy change for the conversion of 36.04 g of ice at -23.0 °C to water at
75.0 °C under a constant pressure of 1 atm.
The specific heats of ice and water are 2.09 J/(°C . g ) and 4.18 J/(°C . g ) respectively.
The heat of fusion of ice is 6.01 kJ/mol.
how do I solve this??
75.0 °C under a constant pressure of 1 atm.
The specific heats of ice and water are 2.09 J/(°C . g ) and 4.18 J/(°C . g ) respectively.
The heat of fusion of ice is 6.01 kJ/mol.
how do I solve this??
Answers
Answered by
DrBob222
q1 = heat to raise T of ice at -2 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q2 = heat to change ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion water.
q3 = heat to raise T of water at zero C to water at 75 C.
Then q total = q1 + q2 + q3
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q2 = heat to change ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion water.
q3 = heat to raise T of water at zero C to water at 75 C.
Then q total = q1 + q2 + q3
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