Answer
(a) $30.4C^{\circ}$
(b) by a factor of two.
Work Step by Step
(a) We know that
$Q=m_{ice}c_{ice}(T_f-T_i)$
$\implies Q_1=(1.1Kg)(2090J/Kg.C^{\circ})(0^{\circ}-(-5C^{\circ}))=11,500J$
and $Q_2=m_{ice}L_f$
$\implies Q_2=(1.1Kg)(33.5\times 10^4J/Kg)=368,500J$
Now $T=\frac{Q_3}{m_{ice}c_w}-0C^{\circ}$
$\implies T=\frac{520,000J-380,000J}{(1.1Kg)(4186J/Kg.C^{\circ})}-0C^{\circ}=30.4C^{\circ}$
(b) We know that the specific heat of water, the specific heat of ice, and the latent heat of fusion are all constant. For the same change of temperature, the heat energy added to the system is directly proportional to the mass of the ice and hence the heat added to the system is doubled and the mass of the ice is also doubled. Thus, the mass of the ice should be increased by a factor of two.
