Answer
$T=100.36^oC$
$m_s=\frac{1}{3}m$
$m_w=\frac{2}{3}m$
Work Step by Step
$mL_f+mc_w(T-0^oC)=mL_v+mc_w(100^oC-T)$
$333\frac{kJ}{kg}+(4.186\frac{kg}{kg^oC})(T)=2260\frac{kJ}{kg}+(4.186\frac{kg}{kg^oC})(100-T)$
$T=100.36^oC$
Assume $m=1kg$ to find ratio of steam to water
$Q=mL_f+mc_w(T-0^oC)=751.6kJ$
$Q=mL_v$
$m_s=\frac{Q}{L_v}=\frac{751.6kJ}{2260\frac{kJ}{kg}}=\frac{1}{3}kg$
$m_w=\frac{2}{3}kg$
