Question

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 1

`m_(Al)=43.71gAl`

Step-by-step explanation:

Hello,

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

`\Delta H_(Al)=-\Delta H_(water)`

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

`m_(Al)Cp_(Al)(T_(eq)-T_(Al))=-m_(water)Cp_(water)(T_(eq)-T_(water))`

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

`m_(Al)=(-m_(water)Cp_(water)(T_(eq)-T_(water)))/(Cp_(Al)(T_(eq)-T_(Al))) \\\\m_(Al)=(-200.0g*4.18(J)/(g^oC)(18.9-15.5)^oC)/(0.897(J)/(g^oC)(18.9-91.4)^oC) \\\\m_(Al)=43.71gAl`

Best regards.