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:
\[q=-753-6030-368\]
\[=-7151J\]
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