Answer: The initial temperature of the iron was [tex]515^0C[/tex]
Explanation:
[tex]heat_{absorbed}=heat_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] .................(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of iron = 360 g
[tex]m_2[/tex] = mass of water = 750 g
[tex]T_{final}[/tex] = final temperature = [tex]46.7^0C[/tex]
[tex]T_1[/tex] = temperature of iron = ?
[tex]T_2[/tex] = temperature of water = [tex]22.5^oC[/tex]
[tex]c_1[/tex] = specific heat of iron = [tex]0.450J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water= [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)][/tex]
[tex]T_i=515^0C[/tex]
Therefore, the initial temperature of the iron was [tex]515^0C[/tex]
