Answer
(a) $2.62KJ$
(b) $30.9KJ$
(c) $0.058Kg;0.68Kg$
Work Step by Step
(a) We know that
$Q_{water}=m_{water}C_{water}(100C^{\circ}-50C^{\circ})$
$\implies Q_{water}=(12.5g)(4186J/Kg.C^{\circ})(100C^{\circ}-50C^{\circ})$
$\implies Q_{water}=2.62KJ$
(b) We know that
$Q_1=m_{steam}L_V$
$Q_1=(12.5g)(22.6\times 10^5J/Kg)$
$Q_1=28.25\times 10^3J$
$Q_2=m_{steam}C_{water}(100C^{\circ}-50C^{\circ})=2.62KJ$
Now $Q=Q_1+Q_2$
$\implies Q=28.25\times 10^3J+2.62\times 10^3J=30.9KJ$
(c) We know that
$m=\frac{Q}{c(T_f-T_i)}$
$\implies m=\frac{2.620\times 10^3J}{3,500J/Kg.K(50.0C^{\circ}-37.0C^{\circ})}=0.058Kg$
Now for the second scenario
$m=\frac{Q}{c(T_f-T_i)}$
We plug in the known values to obtain:
$m=\frac{3.09\times 10^4J}{(3,500J/Kg.K)(50.0C^{\circ}-37.0C^{\circ})}=0.68Kg$
