Question

learn by doing
a 310.0 g piece of aluminum initially at 314°c is dropped into 590 ml of water initially at 22.6°c.
assuming that all heat transfer occurs between the aluminum and the water, calculate the final
temperature.
the specific heat of aluminum is 0.897 j/g°c.
first determine which of the following is the correct equation to use for this calculation.
(0.897 j/g°c)(590 g)(314°c - t_f) = -(4.184 j/g°c)(310.0 g)(22.6°c - t_f)
(0.897 j/g°c)(310.0 g)(314°c - t_f) = (4.184 j/g°c)(590 g)(22.6°c - t_f)
(0.897 j/g°c)(310.0 g)(t_f - 314°c) = -(4.184 j/g°c)(590 g)(t_f - 22.6°c)
(0.897 j/g°c)(310.0 g)(t_f - 314°c) = -(4.184 j/g°c)(590 g)(22.6°c - t_f)
(0.897 j/g°c)(310.0 g)(t_f - 314°c) = (4.184 j/g°c)(590 g)(t_f - 22.6°c)
what is the final temperature of the water and aluminum? (round your answer to the nearest
whole number.)
°c

Explanation:

Step1: Calculate the mass of water

The density of water is \(1\ g/mL\). Given the volume of water \(V = 590\ mL\), using the formula \(m=
ho V\), we get \(m_{water}=590\ g\).

Step2: Use the heat - transfer formula

The heat lost by the aluminum (\(q_{Al}\)) is equal to the heat gained by the water (\(q_{w}\)). The heat - transfer formula is \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat, and \(\Delta T=T_{f}-T_{i}\).
For aluminum: \(q_{Al}=m_{Al}c_{Al}(T_{f}-T_{i,Al})\), where \(m_{Al} = 310.0\ g\), \(c_{Al}=0.897\ J/g^{\circ}C\), \(T_{i,Al}=314^{\circ}C\)
For water: \(q_{w}=m_{w}c_{w}(T_{f}-T_{i,w})\), where \(m_{w} = 590\ g\), \(c_{w}=4.184\ J/g^{\circ}C\), \(T_{i,w}=22.6^{\circ}C\)
Since \(q_{Al}=-q_{w}\) (heat lost by aluminum = heat gained by water), we have \((0.897\ J/g^{\circ}C)(310.0\ g)(T_{f}-314^{\circ}C)=-(4.184\ J/g^{\circ}C)(590\ g)(T_{f}-22.6^{\circ}C)\)

Step3: Expand the equation

$$
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$$

Step4: Combine like terms

$$
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$$

Answer:

\(52^{\circ}C\)