# How can I calculate the specific heat of aluminum?

May 17, 2014

Design and conduct an experiment in which you can calculate the specific heat of aluminum by creating a thermal equilibrium system in which two different with different initial temperatures reach a final temperature that is the same for both.

First examine the design of this experiment.

First heat a 10 gram aluminum metal in beaker of boiling water for at least 10 minutes so that the metal's initial temperature is 100 degrees Celsius.

Using tongs transfer the metal to beaker with 100 grams of water at temperature of 20 degrees Celsius. Measure the final temperature the water. The final temperature is 21.6 Celsius.
The temperature should rise slightly and the metal's temperature should drop dramatically. Why?

First, the water has a higher specific heat so it's temperature will slight compared to a metal which has a specific heat less than one.
Secondly, the mass of the water is 10 times greater.

Now consider the calculation.

$q = m c \left({t}_{f} - {t}_{i}\right)$

q represents heat energy in joules
c represents the specific with units joules/(grams x Celsius)
${t}_{f}$ is final temperature and ${t}_{i}$ is initial temperature.

The q for the metal is negative because it loses heat (exothermic)
The q for the water is positive because it absorbs heat (endothermic)

Since the system will reach thermal equilibrium we will make these equal
- q metal = q water
-(10g)(x)(21.6 C- 100.0 C)= (100 g)(4.184 J/gC)(21.6 C - 20.0 C)

$$             x= 0.90 J/gC


I hope that this example clarifies the question.