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

mass of copper cube = D*V = 8.96 * 1.6^3 = 36.70 g

mass of aluminum cube = D*V = 2.70 * 1.64^3 = 11.909 g

-Qmetals = Qwater

-mcopper*Cpcu*(Tf-Tcu) + -maluminium*Cpal*(Tf-Tal) = mwater*Cpwater*(Tf-Twater)

substitute data

-36.70 *0.385*(Tf-55.3) + -11.909 *0.90*(Tf-55.3) = (100*0.998)*4.184*(Tf-21.5)

(-36.70 *0.385 -11.909 *0.90)* (Tf-55.3)

-24.847Tf + 55.3*24.847 = 417.563Tf - 21.5*417.563

Tf(-417.563--24.847) = - 21.5*417.563-55.3*24.847

Tf = ( - 21.5*417.563-55.3*24.847)/((-417.563--24.847) )

Tf = 26.359°C

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