Answer
(a) $24C^{\circ}$
(b) less than
Work Step by Step
(a) We know that
$T_i=\frac{m_iC_iT_{\circ i}+m_wC_wT_{\circ w}}{m_iC_i+m_wC_w}$
We plug in the known values to obtain:
$T_i=\frac{(0.50Kg)(448J/Kg.K)(723K)+(25Kg)(4186J/Kg.K)(296K)}{(0.50Kg)(448J/Kg.K)+(25Kg)(4186J/Kg.K)}=297K=(297-273)C^{\circ}=24C^{\circ}$
(b) For lead, we can write the equation for the final temperature as follows:
$T_L=\frac{m_LC_LT_{\circ L}+m_wC_wT_{\circ w}}{m_LC_L+m_wC_w}$
We plug in the known values to obtain:
$T_L=\frac{(1.0Kg)(128J/Kg.K)(273K)+(25Kg)(4186J/Kg.K)(296K)}{(1.0Kg)(128J/Kg.K)+(25Kg)(4186J/Kg.K)}$
This simplifies to:
$T_L=296.5K=(296.5-273)C^{\circ}=23.5^{\circ}$
Thus, we can see that the equilibrium temperature is less than that found in part (a).
