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**

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