Answer:
[tex]356^{\circ}C[/tex]
Explanation:
When the hot copper and the liquid water reaches equilibrium, they have the same temperature; when this happens, the amount of heat released by the copper is equal to the amount of heat absorbed by the water:
[tex]-Q_c = Q_w\\-m_c C_c (T_f-T_c) = m_w C_w (T_f-T_w)[/tex]
where
[tex]m_c = 100 g[/tex] is the mass of the copper
C_c=0.385 J/g˚C is the specific heat of copper
T_f=50˚C is the final temperature of both substances
[tex]m_w = 100 g[/tex] is the mass of the water
C_w=4.184 J/g˚C is the specific heat of the water
T_c is the initial temperature of the copper
T_w=20˚C is the initial temperature of the water (room temperature)
Solving for T_c, we find:
[tex]T_c = T_f + \frac{m_w C_w (T_f-T_w)}{m_c C_c}=50^{\circ} +\frac{(100 g)(4.184 J/gC) (50C-20C)}{(100 g)(0.385 J/gC)}=356^{\circ} C[/tex]
