Given:
The mass of the iron is
[tex]m_i=\text{ 352 g}[/tex]
The specific heat of iron is
[tex]c_i=\text{ 0.412 J/g }^{\circ}C[/tex]
The initial temperature of the iron is
[tex]T_i=\text{ 80.2 }^{\circ}C[/tex]
The mass of the water is
[tex]m_w=\text{ 443 g}[/tex]
The specific heat of water is
[tex]c_w=\text{ 4.1813 J/g }^{\circ}C[/tex]
The initial temperature of the water is
[tex]T_w=\text{ 0}^{\circ}C[/tex]
Required: Final equilibrium temperature.
Explanation:
Let the final equilibrium temperature be T.
The final equilibrium temperature can be calculated as
[tex]\begin{gathered} m_ic_i(T-T_i)=m_wc_w(T-T_w) \\ 352\times0.412(T-80.2)=443\times4.1813(T-0) \\ 145.024T-11630.92=1852.32T \\ T=\frac{-11630.92}{1707.29} \\ =\text{ -6.81 }^{\circ}C \end{gathered}[/tex]
