Question

A copper cube measuring 1.58 cm on edge and an aluminum cube measuring 1.68 cm on edge are both heated to 58.7 ∘C and submerged in 100.0 mL of water at 21.9 ∘C. What is the final temperature of the water when equilibrium is reached? (Assume a density of 0.998 g/mL for water.)

Answer 1

A copper cube measuring 1.58 cm on edge and an aluminum cube measuring 1.68 cm on edge are both heated to 56.4 C and submerged in 100.0 mL of water at 21.6 C.

First, we need to calculate the mass of the copper cube and the mass of aluminum cube. Both can be found by using the formula for density:

density = mass / volume

or

mass = (density)(volume)

mass Cu = (8.92 g/cm^3)(1.58 cm)^3 = (8.92 g/cm^3)(3.944312 cm^3)

mass Cu = 35.183263 g

mass Al = (2.70 g/cm^3)(1.68 cm)^3 = (2.70 g/cm^3)(4.741632 cm^3)

mass Al = 12.802406 g

heat lost by both metals = heat gained by water = q

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

heat lost by Al = [(mass Al)(specific heat Al)(final temperature Al - initial temperature Al)

heat gained by H2O = [(mass H2O)(specific heat H2O)(final temperature H2O - initial temperature H2O)]

Note that the final temperature is the same for all.

– q of metals = + q of water

– [(mass Cu)(specific heat Cu)(final temperature Cu – initial temperature Cu) + (mass Al)(specific heat Al)(final temperature Al – initial temperature Al) = [(mass H2O)(specific heat H2O)(final temperature H2O – initial temperature H2O)]

– [(35.183263 g)(0.386 j/g•°C)(final temperature – 56.4°C) + (12.802406 g)(0.900 j/g•°C)(final temperature – 56.4°C) = [(100.0 g)(4.186 j/g•°C)(final temperature - 21.6 °C)]

– [(13.580735 J/°C)(final temperature – 56.4°C) + (11.522165 J/°C)(final temperature – 56.4°C) = [(418.6 J/°C)(final temperature – 21.6°C)]

– [13.580735 J/°C (final temperature) – 765.95345 J + 11.522165 (final temperature) – 649.850 J] = [418.6 J/°C (final temperature) – 9041.76 J]

– 13.580735 J/°C (final temperature) + 765.95345 J – 11.522165 (final temperature) + 649.850 J = 418.6 J/°C (final temperature) – 9041.76 J

9041.76 J + 765.95345 J + 649.850 J = 13.580735 J/°C (final temperature) + 11.522165 J/°C (final temperature) + 418.6 J/°C (final temperature)

10457.663 J = 443.7025 J/°C (final temperature)

final temperature = (10457.663 J) / (443.7025 J/°C)

final temperature = 23.6367°C or 23.6°C rounded to three significant figures