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A slow chemical reaction takes place in a fluid at the constant pressure of 0.1 MPa. The fluid is surrounded by a perfect heat insulator during the reaction which begins at state 1 and ends at state 2. The insulation is then removed and 105 kJ of heat flows to the surroundings as the fluid goes to state 3. The following data are observed for the fluid at states 1, 2, 3.?
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A slow chemical reaction takes place in a fluid at the constant pressu...
Given Information:



  • Constant Pressure: 0.1 MPa

  • Heat Insulator

  • Reaction begins at State 1 and ends at State 2

  • 105 kJ of heat flows to the surroundings during the transition from State 2 to State 3

  • Data for the fluid at States 1, 2, and 3 are known



Solution:


Explanation of States:



  • State 1: Initial State of the Fluid

  • State 2: Final State of the Fluid after the Chemical Reaction

  • State 3: Final State of the Fluid after Heat Flow to Surroundings



Analysis of the Chemical Reaction:



  • As the reaction takes place in a Heat Insulator, there is no change in the Heat Content of the Fluid during the reaction

  • Therefore, the Enthalpy Change during the reaction is equal to the Heat Transfer during the transition from State 2 to State 3

  • Enthalpy Change during the Reaction: ΔH = H2 - H1

  • Heat Transfer during the Transition from State 2 to State 3: Q = H3 - H2

  • Therefore, Q = ΔH = H2 - H1



Determination of Enthalpy Change:



  • Enthalpy Change during the Reaction: ΔH = H2 - H1

  • As the Fluid is at Constant Pressure, the Enthalpy Change can be determined using the following equation:

  • ΔH = CpΔT, where Cp is the Specific Heat Capacity of the Fluid at Constant Pressure and ΔT is the Temperature Change during the Reaction

  • Therefore, ΔH = Cp(T2 - T1)



Determination of Heat Transfer:



  • Heat Transfer during the Transition from State 2 to State 3: Q = H3 - H2

  • As the Fluid is at Constant Pressure, the Heat Transfer can be determined using the following equation:

  • Q = CpΔT, where Cp is the Specific Heat Capacity of the Fluid at Constant Pressure and ΔT is the Temperature Change during the Transition from State 2 to State 3

  • Therefore, Q = Cp(T3 - T2)



Determination of Temperature Changes:



  • Temperature Change during the Reaction: ΔT = T2 - T1 = ΔH/Cp

  • Temperature Change during the Transition from State 2 to State 3: ΔT = T3 - T2 = Q/Cp

Community Answer
A slow chemical reaction takes place in a fluid at the constant pressu...
E2=-29.7KJ , E=-110.7kJ
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A slow chemical reaction takes place in a fluid at the constant pressure of 0.1 MPa. The fluid is surrounded by a perfect heat insulator during the reaction which begins at state 1 and ends at state 2. The insulation is then removed and 105 kJ of heat flows to the surroundings as the fluid goes to state 3. The following data are observed for the fluid at states 1, 2, 3.?
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