Answer
Please see the work below.
Work Step by Step
(a) We know that $Q_A=mC\Delta T$
$\implies Q_A=(1Kg)(2090J/Kg.C^{\circ})(20C^{\circ})=4.18\times 10^4J$
$Q_B=Q_A=mL_f$
$Q_B=4.18\times 10^4J+(1Kg)(33.5\times 10^4J/Kg)=3.77\times 10^5J$
$Q_C=Q_B+mC_{water}\Delta T$
$\implies Q_C=3.77\times 10^5J+(1Kg)(4186J/Kg.C^{\circ})(100C^{\circ})=7.96\times 10^5J$
and $Q_D=Q_C+mL_V$
$Q_D=7.96\times 10^5J+(1Kg)(22.6\times 10^5J)=3.06\times 10^6J$
(b) We know that the slope of the line from point B to point C is given as
$\frac{T_C-T_B}{Q_C-Q_B}=\frac{100C^{\circ}-0C^{\circ}}{7.96\times 10^5J-3.77\times 10^5J}=2.39\times 10^{-4}C^{\circ}/J$
Now $\frac{1}{C}=\frac{1}{4186J/Kg.C^{\circ}}=2.39\times 10^{-4}C^{\circ}/J=slope \space of \space line \space from \space B\space to \space C$
