Asked by Johnny
Lithium metal is a highly reactive metal that oxidizes instantly in water or air. Given the data below, calculate the energy required to heat 10.0 g of Li from 150.0 °C to 200.0 °C.
Molar heat capacity (solid) = 3.58 J/°C • mol
Molar heat capacity (liquid) = 4.379 J/°C • mol
Melting point = 180.5 °C
Boiling point = 1342 °C
Heat of fusion = 3.00 kJ/mol
Heat of vaporization = 147.1 kJ/mol
Molar heat capacity (solid) = 3.58 J/°C • mol
Molar heat capacity (liquid) = 4.379 J/°C • mol
Melting point = 180.5 °C
Boiling point = 1342 °C
Heat of fusion = 3.00 kJ/mol
Heat of vaporization = 147.1 kJ/mol
Answers
Answered by
DrBob222
The melting point of Li is 180.5.
q1 = heat to raise T of solid from 150 to 181.
q1 = mass Li x specific heat solid Li x (Tfinal-Tinitial). Note: Tf is 181; Ti is 150.
q2 = heat to melt Li at 180.5.
q2 = mass Li x heat fusion.
q3 = heat to raise T of liquid Li from 180.5 to 200.0
q3 = mass Li x specific heat liquid Li x (Tfinal-Tinitial). Note: Tf = 200; Ti = 180.5.
Total Q = q1 + q2 + q3.
q1 = heat to raise T of solid from 150 to 181.
q1 = mass Li x specific heat solid Li x (Tfinal-Tinitial). Note: Tf is 181; Ti is 150.
q2 = heat to melt Li at 180.5.
q2 = mass Li x heat fusion.
q3 = heat to raise T of liquid Li from 180.5 to 200.0
q3 = mass Li x specific heat liquid Li x (Tfinal-Tinitial). Note: Tf = 200; Ti = 180.5.
Total Q = q1 + q2 + q3.
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