A) 5649,369
B) 5544,324
C) 5278,342
D) 3425,425
Correct Answer: A
Solution :
[a] 18gm of water at \[100{}^\circ C\] \[10gm\] of Cu at \[25{}^\circ C\] is added. \[{{q}_{p}}={{C}_{p,m}}dt\] \[=75.32\times \frac{J}{K\,mol}\times \frac{18g}{18g/mol}\left( 373-298 \right)K\] \[=75.32\frac{J}{K}\times 75K\] \[=5.649\times {{10}^{3}}J\] If now \[10g\] of copper is added \[{{C}_{p,m}}=24.47J/mol\,K\] Amount of heat gained by Cu \[=24.47\times \frac{J}{K\,mol}\times \frac{10g}{63g/mol}\left( 373-298 \right)K\] \[=291.3\text{ }J\] Heat lost by water \[=291.30\text{ }J\] \[-291.30J=75.32\frac{J}{K}\times ({{T}_{2}}-373K)\] \[\Rightarrow -3.947\text{ }K={{T}_{2}}-373K\] \[\Rightarrow {{T}_{2}}=369.05K\]You need to login to perform this action.
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