Answer: a) [tex]38.3^0C[/tex]
Explanation:
As we know that,
[tex]q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat released = -600.2 J
[tex]m[/tex] = mass of iron rod =100 g
[tex]T_{final}[/tex] = final temperature =[tex]25.^0C[/tex]
[tex]T_{initial}[/tex] = initial temperature = ?
[tex]c[/tex] = specific heat of iron = [tex]0.452J/g^0C[/tex]
Now put all the given values in equation, we get
[tex]-600.2J=100\times 0.452J/g^0C\times (25.0-T_{initial})^0C][/tex]
[tex]T_{initial}=38.3^0C[/tex]
Therefore, the initial temperature of iron rod was [tex]38.3^0C[/tex]