Given:
Quantity of heat absorbed (Q)=54 calories= 226 J
m=58.3g=0.0583Kg
Tinitial:12°c=285.15 K
Tfinal:42°C=315.15K
Now we know that
Q= mc∆T
Where Q is the heat absorbed measured in J
m is the mass of the substance measured in Kg
c is the specific heat measured in
J/Kg.K
∆T is the change in temperature measured in K
Substituting the given values in the above formula we get
226=0.0583 x c x(315.15-285.15)
c=129.22 J/Kg.K
Thus the specific heat of lead is
129.22 J/Kg.K .
