Question

If 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C, what is the specific heat capacity of the metal?

Answer 1

The specific heat capacity is an amount of heat required for 1 kg of material to rise its temperature on 1 C degree:

$C=\frac{Q}{m\delta t}=\frac{Q}{m(t_2-t_1)}=\frac{1495J}{0.315kg*11.0^0C}=431\frac{J}{kg^0C};$

On the other hand, we can even try to say something about this metal:

$C=\frac{3R}{\mu};$

$\mu=\frac{3R}{C}=\frac{3*8.31\frac{J}{mol*K}}{431\frac{J}{kg*K}}=0.0578kg/mol.$

This metal is most probably something like Fe, Co or Ni.