Question

a hot lump of 34.6 g of aluminum at an initial temperature of 64.9°c is placed in 50.0 ml h₂o initially at 25.0°c and allowed to reach thermal equilibrium. what is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 j/(g·°c)? assume no heat is lost to surroundings.

Explanation:

Step1: Calculate the mass of water

The density of water is \(1\ g/mL\). Given \(V = 50.0\ mL\), using \(m=
ho V\), we have \(m_{water}=1\ g/mL\times50.0\ mL = 50.0\ g\). The specific heat of water \(c_{water}=4.184\ J/(g\cdot^{\circ}C)\).

Step2: Set up the heat - transfer equation

Since \(q = mc\Delta T\) and \(q_{aluminum}=-q_{water}\) (no heat loss to surroundings). Let the final temperature be \(T_f\).
\(q_{aluminum}=m_{aluminum}c_{aluminum}(T_f - T_{i,aluminum})\) and \(q_{water}=m_{water}c_{water}(T_f - T_{i,water})\)
So, \(m_{aluminum}c_{aluminum}(T_f - T_{i,aluminum})=-m_{water}c_{water}(T_f - T_{i,water})\)
Substitute \(m_{aluminum}=34.6\ g\), \(c_{aluminum}=0.903\ J/(g\cdot^{\circ}C)\), \(T_{i,aluminum}=64.9^{\circ}C\), \(m_{water}=50.0\ g\), \(c_{water}=4.184\ J/(g\cdot^{\circ}C)\) and \(T_{i,water}=25.0^{\circ}C\) into the equation:

$$
LATEXBLOCK0
$$

Answer:

\(30.2^{\circ}C\)