Question

A total of 54.0 calories of heat are absorbed as 58.3 g of lead is heated from 12.0° C to 42.0° C. What is the specific heat of lead?

Answer 1

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 .