The following information is given for aluminum at 1 atm: T=2467.00°C Tm=660.00°C Specific heat solid = 0.9000 Specific heat liquid = 1.088 A 42.30 g sample of solid aluminum is initially at 640.00°C. If the sample is heated at constant pressure (P = 1 atm), of the sample to 1030 00°C g. °C AHvap (2467.00° C) = 10530 J/g AHfus (660.00°C) = 398.4 J/g kJ of heat are needed to raise the temperature

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The following information is given for aluminum at 1 atm:

- Boiling point (Tb) = 2467.00°C
- Melting point (Tm) = 660.00°C

- Specific heat of solid aluminum = 0.9000 J/(g·°C)
- Specific heat of liquid aluminum = 1.088 J/(g·°C)

- Enthalpy of vaporization at 2467.00°C (ΔHvap) = 10530 J/g
- Enthalpy of fusion at 660.00°C (ΔHfus) = 398.4 J/g

A 42.30 g sample of solid aluminum is initially at 640.00°C. If the sample is heated at constant pressure (P = 1 atm), how many kJ of heat are needed to raise the temperature of the sample to 1030.00°C? 

(Note: The solution involves calculating the heat required to first raise the sample to its melting point, then the heat required for the phase change, and finally the heat needed to increase the temperature of the liquid.)
Transcribed Image Text:The following information is given for aluminum at 1 atm: - Boiling point (Tb) = 2467.00°C - Melting point (Tm) = 660.00°C - Specific heat of solid aluminum = 0.9000 J/(g·°C) - Specific heat of liquid aluminum = 1.088 J/(g·°C) - Enthalpy of vaporization at 2467.00°C (ΔHvap) = 10530 J/g - Enthalpy of fusion at 660.00°C (ΔHfus) = 398.4 J/g A 42.30 g sample of solid aluminum is initially at 640.00°C. If the sample is heated at constant pressure (P = 1 atm), how many kJ of heat are needed to raise the temperature of the sample to 1030.00°C? (Note: The solution involves calculating the heat required to first raise the sample to its melting point, then the heat required for the phase change, and finally the heat needed to increase the temperature of the liquid.)
Expert Solution
Step 1

Given,

The melting point of Al = 660.0 °C

Initial temperature = 640.0°C

Final temperature =  1030.00 °C

Mass of sample = 42.30 g

Heat requires can be calculated by 

Q = m.S.TWhere S = specific heat of sample m = mass of sample

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