Answer to Question #84300 in General Chemistry for Stephanie Ezman

Question #84300
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.)
1
Expert's answer
2019-01-18T04:22:25-0500

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

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