Answer: [tex]22.8^0C[/tex]
Explanation:-
[tex]Q_{absorbed}=Q_{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\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]
[tex]m_2[/tex] = mass of water = 21.9 kg = 21900 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of iron horseshoe = [tex]600^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]21.8^oC[/tex]
[tex]c_1[/tex] = specific heat of iron horseshoe = [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]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]
[tex]350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)][/tex]
[tex]T_{final}=22.8^0C[/tex]
Therefore, the final equilibrium temperature is [tex]22.8^0C[/tex].
