Step 1: Calculate the average specific heat of the metals:

$\overline{){{\mathbf{C}}}_{{\mathbf{ave}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{Al}}}{\mathbf{Mass}}{\mathbf{}}{{\mathbf{ratio}}}_{{\mathbf{Al}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{Fe}}}{\mathbf{Mass}}{\mathbf{}}{{\mathbf{ratio}}}_{{\mathbf{Fe}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{C}}_{\mathbf{ave}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{89}\frac{\mathbf{J}}{\mathbf{g}\mathbf{\xb7}\mathbf{\xb0}\mathbf{C}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{}\mathbf{Fe}}{\mathbf{15}\mathbf{}\mathbf{g}\mathbf{}\mathbf{metal}}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{45}\frac{\mathbf{J}}{\mathbf{g}\mathbf{\xb7}\mathbf{\xb0}\mathbf{C}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{}\mathbf{Al}}{\mathbf{15}\mathbf{}\mathbf{g}\mathbf{}\mathbf{metal}}\mathbf{\right)}$

**C _{ave }= 0.60 J/g•°C**

A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 J/°C ? g) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 J/°C ? g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 97.3 g water at 22.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

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