Question

The specific heat of a certain type of metal is 0.128 j/(g·°c). what is the final temperature if 305 j of heat is added to 64.6 g of this metal initially at 20.0 °c?

Answer 1

When we add a certain amount of heat Q to a substance, the temperature of the substance increases by a $\Delta T$ given by
$Q=m C_s \Delta T$
where m is the mass of the substance and Cs is the  specific heat capacity of the substance.

By re-arranging the formula, we find
$\Delta T = \frac{Q}{m C_s}= \frac{305.0 J}{(64.6 g)(0.128 J/gC)}=36.9 ^{\circ}C$

So, since the initial temperature of the metal is Ti=20 C, the final temperature is
$T_f = T_i + 36.9 ^{\circ} C=20.0^{\circ} + 36.9^{\circ} C=56.9^{\circ}C$