Answer to Question #84300 in General Chemistry for Stephanie Ezman

Question

A copper cube measuring 1.56 cm on edge and an aluminum cube measuring 1.66 cm on edge are both heated to 57.6 ∘C and submerged in 100.0 mL of water at 21.8 ∘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

Specific heat od aluminum = 0.900 J/g*C

Specific heat of copper = 0.386 J/g*C

Specific heat of water = 4.186 J/g*C

density of aluminum = 2.70g/cm³

density of copper = 8.94 g/cm³

m = d*V

Volume of a cube V=a³

At equilibrium: heat lost my both metals = heat gained by water

-[c(Al)*m(Al)*(T₂ - 57.6) + c(Cu)*m(Cu)*(T₂ - 57.6)] = c(water)* m(water)*(T₂ - 21.8)

- [c(Al)*d(Al)*V(Al)(T₂ - 57.6) + c(Cu)*d(Cu)*V(Cu)*(T₂ - 57.6)] = c(water)* d(water)*V(water)*(T₂ - 21.8)

-[0.900*2.7*(1.66)³*(T₂ - 57.6) + 0.386*8.94*(1.56)³*(T₂ - 57.6)] = 4.186*0.998*100.0*(T₂ -21.8)

-[11.16*(T₂ - 57.6) + 13.10*(T₂ - 57.6)] = 417.76*(T₂ - 21.8)

(57.6 - T₂) *24.26 = 417.76*(T₂ - 21.8)

57.6 - T₂ = 17.22 (T₂ - 21.8)

57.6 - T₂ = 17.22 *T₂ - 375.4

433 = 18.22*T₂

T₂ = 23.77 C

Answer: 23.77 C