Respuesta :
Answer
117.64 g
Procedure
Upon inspection, one of the given heat capacities is in J/°C mol; we will convert it to J/g°C
[tex]24.03\frac{\text{J}}{mol\text{ Al }\degree\text{C}}\frac{1\text{ mol Al}}{26.9815\text{ g Al}}=0.889498\text{ }\frac{\text{J}}{\text{ g}\degree\text{C}}[/tex]To solve this problem, we will assume that the heat released by the aluminum piece is the same as the heat absorbed by the water.
We have the specific heat formula
[tex]Q=mc\Delta T[/tex]Where
Q=heat energy
m=mass
c=specific heat capacity
Δ T=change in temperature
Then we will compare both equations and consider the variables, with subindex 1 being for water and 2 for aluminum
[tex]m_1c_1\Delta T_1=m_2c_2\Delta T_2[/tex]Solving for m1, we have
[tex]m_1=\frac{m_2c_2\Delta T_2}{c_1\Delta T_1}=\frac{-25\text{ g }0.8894\text{ J/ g}\degree\text{C}(24.9-82.4)\degree\text{C}}{4.18\text{ J/ g}\degree\text{C}(24.9-22.3)\text{ }\degree\text{C}}=117.64\text{ g}[/tex][tex]m_1=m_2c_2\Delta T_2=25\text{ g }0.8894\text{ J/ g}\degree\text{C}(24.9-82.4)\degree\text{C}=-1278.51\text{ g}[/tex][tex]m_2=\frac{Q_2}{c_1\Delta T_1}=\frac{1278.51}{4.18\text{ J/ g}\degree\text{C }(24.9-22.3)}=117.64\text{ g}[/tex]