Answer
$ -21.2 J/K$
Work Step by Step
The process of cooling the water goes by
$\int ^{339.67} _{353.15} \frac{cm'}{T} dT $
$ = cm' [ln T]^{339.67} _{353.15}$
$ = cm' [ln \frac{339.67}{353.15}]$
Where $ c =4190 J/Kg.^oC$
and $ m' = 0.130 kg$
$ = (4190 J/Kg.^oC)(0.130 kg) [ln \frac{339.67}{353.15}]$
$= -21.2 J/K$
The negative indicates that heat is released out of the system.
