Question

a 56.8 g sample of aluminum is put into a calorimeter (see sketch at right) that contains 150.0 g of water. the aluminum sample starts off at 85.2 °c and the temperature of the water starts off at 20.0 °c. when the temperature of the water stops changing it’s 24.1 °c. the pressure remains constant at 1 atm. calculate the specific heat capacity of aluminum according to this experiment. be sure your answer is rounded to 2 significant digits.

Explanation:

Step1: Define heat transfer balance

Heat lost by Al = Heat gained by water: $m_{Al}c_{Al}\Delta T_{Al} = m_{water}c_{water}\Delta T_{water}$

Step2: List known values

$m_{Al}=56.8g$, $\Delta T_{Al}=85.2-24.1=61.1^\circ C$; $m_{water}=150.0g$, $c_{water}=4.184J/(g\cdot^\circ C)$, $\Delta T_{water}=24.1-20.0=4.1^\circ C$

Step3: Solve for $c_{Al}$

$c_{Al}=\frac{m_{water}c_{water}\Delta T_{water}}{m_{Al}\Delta T_{Al}}=\frac{150.0\times4.184\times4.1}{56.8\times61.1}$

Step4: Calculate and round

$c_{Al}\approx\frac{150\times4.184\times4.1}{56.8\times61.1}\approx\frac{2572.44}{3470.48}\approx0.74J/(g\cdot^\circ C)$

Answer:

0.74 J/(g·°C)