Question

If 1495 j of heat is needed to raise the temperature of a 361 g sample of a metal from 55.0°c to 66.0°c, what is the specific heat capacity of the metal?

Answer 1

When a certain amount of heat Q is absorbed by a substance, its temperature increases by $\Delta T$ following the relationship
$Q= m C_s \Delta T$
where m is the mass of the substance and $C_s$ is the specific heat of the substance.

For the metal in our problem, the mass of the sample is m=361 g, the amount of heat is Q=1495 J, and the temperature difference is 
$\Delta T = 66C-55C=11^{\circ} C$
so by re-arranging the previous formula and by substituting the numbers we can find the specific heat of the metal:
$C_s = \frac{Q}{m \Delta T}= \frac{1495 J}{(361 g)(11 ^{\circ}C)}=0.38 Jg^{-1} C^{-1}$