Answer
(a) $8.7\times 10^{-4}C^{\circ}/J$
(b) $4.34\times 10^{-4}C^{\circ}/J$
Work Step by Step
(a) We know that
$Q_B=mc_{ice}\Delta T$
$Q_B=(0.55Kg)(2090J/Kg.C^{\circ})(0C^{\circ}-(-20C^{\circ}))=2.3\times 10^4J$
Now $slope=\frac{T_B-T_A}{Q_B-Q_A}$
$\implies slope=\frac{0C^{\circ}-(-20C^{\circ})}{2.3\times 10^4J-0J}=8.7\times 10^{-4}C^{\circ}/J$
$\frac{1}{mC_{ice}}=\frac{1}{(0.55Kg)(2090J/Kg.C^{\circ})}=8.7\times 10^{-4}C^{\circ}/J$
(b) We know that
$slope=\frac{T_D-T_C}{Q_D-Q_C}$
We plug in the known values to obtain:
$slope=\frac{20C^{\circ}}{mC_{water}\Delta}$
We plug in the known values to obtain:
$slope=\frac{20C^{\circ}}{(0.55Kg)(4186J/Kg.C^{\circ})(20C^{\circ})}$
$\implies \frac{1}{mC_{water}}=\frac{1}{(0.55Kg)(4186J/Kg.C^{\circ})}=4.34\times 10^{-4}C^{\circ}/J$
