question_answer

What is the amount of heat (in Joules) absorbed by 18 g of water initially at room temperature heated to \[100{}^\circ C\]? If 10 g of Cu is added to this water, than decrease in temperature (in Kelvin) of water was found to be? C (p,m) for water\[75.32\text{ }J/mol\text{ }K\]; C (p,m) for \[Cu=24.47\text{ }J/mol\text{ }K.\]

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 \[=291.30\text{ }J\] \[-291.30J=75.32\frac{J}{K}\times ({{T}_{2}}-373K)\] \[\Rightarrow -3.947\text{ }K={{T}_{2}}-373K\] \[\Rightarrow {{T}_{2}}=369.05K\]