The lead and the water reach thermal equilibrium when the net flow of heat from the lead to the water becomes zero, so when the heat released by the lead [tex]Q_L[/tex] is equal to the heat absorbed by the water [tex]Q_w[/tex]:
[tex]-Q_L = Q_w[/tex] (1)
where the negative sign means the heat of the lead is released, not absorbed.
The equation that gives the heat absorbed/released is
[tex]Q=m C \Delta T[/tex] (2)
where
m is the mass of the substance
C is its specific heat capacity
[tex]\Delta T[/tex] is the variation of temperature.
If we rewrite (1) by using (2), we get
[tex]-m_L C_L \Delta T_L = m_w C_w \Delta T_w[/tex] (3)
where the label L refers to the lead and the label W refers to the water.
The total mass of the lead is:
[tex]m_L = 3 \cdot 1.00 \cdot 10^2 g =300 g[/tex]
while its change of temperature is
[tex]\Delta T_L = T_f - T_i = 45.0^{\circ} C - 100^{\circ}C = -55^{\circ} C[/tex]
The mass of the water is
[tex]m=2.00 \cdot 10^2 g = 200 g[/tex]
its specific heat capacity is
[tex]C_w = 4.18 J/(g ^{\circ}C)[/tex]
while its change of temperature is
[tex]\Delta T_w = T_f - T_i = 45.0^{\circ} C- 35.0^{\circ} C= 10^{\circ}C[/tex]
If we put all this data inside (3), we can calculate the specific heat capacity of lead:
[tex]C_L = - \frac{m_w C_w \Delta T_w}{m_L C_L}=- \frac{(200 g)(4.18 J/g^{\circ}C)(10^{\circ}C)}{(300 g)(-55^{\circ}C)}= 0.50 J/(g^{\circ}C) [/tex]
