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A chunk of aluminum at 91.4°C was added to 200.0 g of water at 15.5°C. The specific heat of aluminum is 0.897 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.9°C. Assuming no heat was lost to the surroundings, what was the mass of aluminum added?

A. 243 g
B. 34.7 g
C. 41.7 g
D. 43.7 g

Answer :

sebassandin

Answer:

[tex]m_{Al}=43.71gAl[/tex]

Explanation:

Hello,

In this case, since the heat lost by the aluminium is gained by the water, the following equality is achieved:

[tex]\Delta H_{Al}=-\Delta H_{water}[/tex]

Which in terms of masses, heat capacities and temperatures is:

[tex]m_{Al}Cp_{Al}(T_{eq}-T_{Al})=-m_{water}Cp_{water}(T_{eq}-T_{water})[/tex]

Thus, by solving for the mass of aluminium, we obtain:

[tex]m_{Al}=\frac{-m_{water}Cp_{water}(T_{eq}-T_{water})}{Cp_{Al}(T_{eq}-T_{Al})} \\\\m_{Al}=\frac{-200.0g*4.18\frac{J}{g^oC}(18.9-15.5)^oC}{0.897\frac{J}{g^oC}(18.9-91.4)^oC} \\\\m_{Al}=43.71gAl[/tex]

Best regards.

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