Answer : The specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]
Explanation :
Formula used:
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = 3840 J
m = mass of lead = 3 kg = 3000 g
c = specific heat capacity of lead = ?
[tex]T_1[/tex] = initial temperature = [tex]40^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]50^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]3840J=3000g\times c\times (50-40)^oC[/tex]
[tex]c=0.128J/g^oC[/tex]
Therefore, the specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]