What is the final temperature of 30 grams of aluminum at 95 degrees Celsius added to 100 grams of water at 25 degreees celsius?

Answer 1

To calculate one should use:

[(mass Al).(specific heat Al).( Tf - TiAluminum)] + [(mass H2O).(specific heat H2O).(Tf - Tiwater)] = 0 (Law of conservation of energy or heat)

Tf is the final temperature in ºC (because there is a change in temperature, it does not have to be converted to Kelvin)

Ti is the initial temperature in ºC

Table values:

specific heat of water = 4.18 J/K.g

specific heat of Aluminum = 24.200 J/mol·K = 24.200(J/mol.K) / 26.98(g/mol) = 0.8970 J/K.g

So, solve this:

30*0.8970*(Tf - 95ºC) + 100*4.18*(Tf - 25) = 0

Tf = 29.34 ºC of both water and the immersed aluminum and supposed no water is evaporated.