The following information is given for bismuth at 1 atm: boiling point = 1627 °C melting point = 271 °C specific heat solid = 0.126 J/g°C specific heat liquid = 0.151 J/g°C AHvap(1627 °C) = 172.0 kJ/mol AHfus(271 °C) = 11.00 kJ/mol What is AH in kJ for the process of freezing a 29.1 g sample of liquid bismuth at its normal melting point of 271 °C? kJ

Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 9E: If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?
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**Bismuth Properties at 1 atm:**

- **Boiling Point:** 1627 °C
- **Melting Point:** 271 °C
- **Specific Heat (Solid):** 0.126 J/g°C
- **Specific Heat (Liquid):** 0.151 J/g°C

**Enthalpy Changes:**

- **ΔH<sub>vap</sub> (1627 °C):** 172.0 kJ/mol
- **ΔH<sub>fus</sub> (271 °C):** 11.00 kJ/mol

**Problem Statement:**

Calculate the ΔH (in kJ) for the process of freezing a 29.1 g sample of liquid bismuth at its normal melting point of 271 °C.

\[ \boxed{\phantom{answer\ in\ kJ}} \]
Transcribed Image Text:**Bismuth Properties at 1 atm:** - **Boiling Point:** 1627 °C - **Melting Point:** 271 °C - **Specific Heat (Solid):** 0.126 J/g°C - **Specific Heat (Liquid):** 0.151 J/g°C **Enthalpy Changes:** - **ΔH<sub>vap</sub> (1627 °C):** 172.0 kJ/mol - **ΔH<sub>fus</sub> (271 °C):** 11.00 kJ/mol **Problem Statement:** Calculate the ΔH (in kJ) for the process of freezing a 29.1 g sample of liquid bismuth at its normal melting point of 271 °C. \[ \boxed{\phantom{answer\ in\ kJ}} \]
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