Answer:
46.9 C
Explanation:
The heat released by the gold bar is equal to the heat absorbed by the water:
[tex]m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)[/tex]
where:
[tex]m_g = 3.0 kg[/tex] is the mass of the gold bar
[tex]C_g=129 J/kg C[/tex] is the specific heat of gold
[tex]T_g=99 C[/tex] is the initial temperature of the gold bar
[tex]m_w = 0.22 kg[/tex] is the mass of the water
[tex]C_w=4186 J/kg C[/tex] is the specific heat of water
[tex]T_w=25 C[/tex] is the initial temperature of the water
[tex]T_f[/tex] is the final temperature of both gold and water at equilibrium
We can re-arrange the formula and solve for T_f, so we find:
[tex]m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C[/tex]
