Answer:
103.8 g
Explanation:
When the hot piece of copper is placed in the water at lower temperature, the piece of copper gives off thermal energy to the water; as a result, the temperature of the copper decreases while the temperature of the water increases, until they both reach the equilibrium temperature.
The heat given off by the piece of copper is equal to the heat absorbed by the water, so we can write:
[tex]Q_c=Q_w[/tex]
where:
[tex]Q_w=m_w C_w \Delta T_w[/tex] is the heat absorbed by the water, where
[tex]m_w = 96.2 g[/tex] is the mass of water
[tex]C_w=4.186 J/gC[/tex] is the specific heat of water
[tex]\Delta T_w=5.1C[/tex] is the rise in temperature of the water
Solving,
[tex]Q_w=(96.2)(4.186)(5.1)=2053.7 J[/tex]
[tex]Q_c=m_c C_c (T_c-T)[/tex] is the heat released by the copper, where
[tex]m_c[/tex] is the mass of copper
[tex]C_c=0.385 J/gC[/tex] is the specific heat of copper
[tex]T_c=73.6C[/tex] is the initial temperature of copper
[tex]T=17.1C+5.1C=22.2 C[/tex] is the equilibrium temperature
Solving for the mass,
[tex]m_c=\frac{Q_c}{C_c(T_c-T)}=\frac{2053.7}{(0.385)(73.6-22.2)}=103.8 g[/tex]