Question

Consider the following properties of a particular substance:
melting point: -7.2 degrees C
boiling point: 58.8 degrees C
heat of fusion: 10.571 kJ/mol
specific heat of the solid: 87.4 J/mol degrees C
specific heat of the gas: 40.0 J/mol degrees C
specific heat of the liquid: 151.4 J/mol degrees C
heat of vaporization: 29.96 kJ/mol

A 1.54 mol sample of this substance was heated from -40.0 degrees C to 80.0 degrees C at a constant rate of 623 J/min.

a) How much heat (in kJ) was absorbed in the process?

b) How many minutes does it take to reach room temperature (25 degrees C)?

Answer 1

heat from-40C to -7.2C: =1.54*87.4*32.8 joules
heat to melt at -7.2= 1.54*10.571 *1000 joules
heat from-7.2 to 58.8;= 1.54*151.4 Joules
Heat to boil at 58.8;= 1.54*29.96 *1000 joules
heat from 58.8 to 80= ....

then add the heats.