## Chemistry 9th Edition

$T_f=56.5^{\circ}C$
As given that $5.25cm=10.0^{\circ}F$ For $1cm=\frac{10.0^{\circ}F}{5.25cm}$ Now for $18.5cm$ $18.5cm=(18.5cm)\frac{10.0^{\circ}F}{5.25cm}=35.2^{\circ}F$ Thus $T_f=98.6+35.2=133.8^{\circ}F$ Finally, we convert this final temperature into centigrades $T_f=\frac{5}{9}(133.8-32)=56.5^{\circ}C$