[tex]Q=mc \Delta T \\\\
Q=168,2J\\
m=4g\\
c=4.184\frac{J}{g^{o}C}\\\\
\Delta T=\frac{Q}{mc}=\frac{167,2J}{4g*4,184\frac{J}{g^{o}C}}=9,99^{o}C\\\\
\Delta T = T_{f}-T_{i}\\\\
9,99^{o}C=T_{f}-10^{o}C\\\\
T_{f}=9,99^{o}C+10^{o}C=19,99^{o}C\approx20^{o}C[/tex]
