Question

A copper cube measuring 1.60 cm on edge and an aluminum cube measuring 1.64 cm on edge are both heated to 55.3 ∘C and submerged in 100.0 mL of water at 21.5 ∘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

volume of cube(V) = a^3

   a = edge length

Volume of Cu cube = 1.6^3 = 4.096 cm^3

mass of Cu cube = density*Volume

density of Cu = 8.96 g/cc

       = 8.96*4.096

       = 36.7 g

Volume of Al cube = 1.64^3 = 4.411 cm^3

mass of Al cube = density*Volume

density of Al = 2.7 g/cc

       = 2.7*4.411

       = 11.91 g

heat lost by metal = heat gained by water + vessel

    mass of Cu*sCu*DT + mass of Al*Sal*DT = mass of water*swater*DT

   36.7*0.385*( 55.3-x)+11.91*0.902*(55.3-x) = 100*0.998+4.184*(x-21.5)

x = final temeperature of system = 50 c