Answer
(a) $3.6C^{\circ}$
(b) No ice is present
Work Step by Step
(a) We know that
$Q=mc\Delta T$
For copper
$Q_{cu}=m_{cu}C_{cu}(T_f-T_i)_{cu}$
$Q_{cu}=(48g(\frac{1Kg}{10^3g}))(387J/KgC^{\circ})(T_f-(-12C^{\circ})_{cu})$
For water
$Q_{w}=m_wc_w(T_i-T_f)w$
$Q_{w}=(110(\frac{1Kg}{10^3g}))(4186J/KgC^{\circ})(4.1C^{\circ}-T_f)_w$
For aluminum
$Q_{Al}=(75g(\frac{1Kg}{10^3g}))(900J/KgC^{\circ})(4.1C^{\circ}-T_f)Al$
Now $Q_{cu}=Q_w+Q_{Al}$
We plug in the known values to obtain:
$(48g(\frac{1Kg}{10^3g}))(387J/KgC^{\circ})(T_f-(-12C^{\circ})_{cu})=(110(\frac{1Kg}{10^3g}))(4186J/KgC^{\circ})(4.1C^{\circ}-T_f)_w+(75g(\frac{1Kg}{10^3g}))(900J/KgC^{\circ})(4.1C^{\circ}-T_f)Al$
This simplifies to:
$T_f=3.6C^{\circ}$
(b) We know that the final equilibrium temperature is $3.6C^{\circ}$ and it is greater than $0C^{\circ}$; therefore, there is no ice when the system reaches equilibrium.
