question_answer 23)

Calculate the enthalpy change on freezing of 1 mole of water at\[10{}^\circ C\]to ice at\[-10{}^\circ C\]. \[{{\Delta }_{f}}H=6.03kJ\,mo{{l}^{-1}}at{{0}^{{}^\circ }}C.\] \[{{C}_{p}}[{{H}_{2}}O(l)]=75.3\,\,J\,mo{{l}^{-1}}{{K}^{-1}}\] \[{{C}_{p}}[{{H}_{2}}O(s)]=36.8\,\,J\,mo{{l}^{-1}}{{K}^{-1}}\]  

Answer:

Step 1: Heat released in step 1 \[q=-n{{C}_{p}}\Delta T\] Step 2: Heat released in freezing process \[q=-n{{\Delta }_{freezing}}H\] \[=-1\times 6.03=-6.03kJ\] \[=-6030J\] Step 3: Heat released in the cooling ice \[q=-n{{C}_{p}}\Delta T\] \[=-1\times 6.03=-6.03kJ\] \[=-6030J\] Total heat released: \[q=-753-6030-368\] \[=-7151J\]