Question

A copper cube measuring 1.53 cm on edge and an aluminum cube measuring 1.69 cm on edge are both heated to 57.1 ∘C and submerged in 100.0 mL of water at 21.6 ∘C .

What is the final temperature of the water when equilibrium is reached? (Assume a density of 0.998 gmL−1 for water.)

Answer 1

mass = (density)(volume)

mass Cu = (8.92 g/cm^3)(1.53 cm)^3 = (8.92 g/cm^3)(3.581577 cm^3)

mass Cu = 31.95 g

mass Al = (2.70 g/cm^3)(1.69 cm)^3 = (2.70 g/cm^3)(4.826809 cm^3)

mass Al = 13.03 g

heat lost by Copper and aluminum = heat gained by water = q

heat lost by Cu = – [(mass of Cu)(specific heat of Cu)(final temperature Cu - initial temperature Cu)

heat lost by Al = [(m)(Sh Al)(Tf of Al - Ti Al)

heat gained by H2O = [(m H2O)(Sh H2O)(Tf H2O - Ti H2O)]

the final temperature is the same for all.

– [(31.95 g)(0.386 j/g•°C)(Tf – 57.1°C) + (13.03 g)(0.900 j/g•°C)(Tf – 57.1°C) = [(99.8 g)(4.186 j/g•°C)(Tf - 21.6 °C)]

– [(12.3327 J/°C)(Tf – 57.1°C) + (11.727 J/°C)(Tf – 57.1°C) = [(417.76 J/°C)(Tf – 21.6°C)]

704.197 + 669.6117 -24.0597 Tf = 417.76 Tf - 9020.16

final temperature = (10393.9687 J) / (441.8197 J/°C)

final temperature = 23.52536°C or 23.5°C