Question

A 53.71-g cube of aluminum was heated to 97.33 °C and transferred to a beaker containing 82.15 g of SAE 20W/50 oil at 41.3 °C. Given that the specific heat capacity of the oil is 2.316 J g⁻¹ K ⁻¹, the molar heat capacity of aluminum is 24.55 J mol⁻¹ K ⁻¹, and ignoring heat losses from the assembly, what is the final temperature of the system?

Answer 1

m = 53.71 g of alumnium

Tal = 97.33 C

m = 82.15 g of oil

T = 41.3°C

Cpoil =2.316 J

Cpal = 24.55 J/mol --> 24.55 /26.981539 = 0.91 J/gC

findfinal T

heat balance:

-Qlost = Qgain

Qaluminium = Qoil

Qal = mass of Al- Cp Al * (Tf-Tal)

Qal = mass of Oil - Cp Oil* (Tf-Toil)

substitute

-mass of Al- Cp Al * (Tf-Tal) = mass of Oil - Cp Oil* (Tf-Toil)

- 53.71 * 0.91*(Tf-97.33) = 82.15*2.316 *(Tf-41.3)

solve for Tf

- 53.71 * 0.91/(82.15*2.316) * (Tf- 97.33) = Tf-41.3

-0.256 * ( (Tf- 97.33) ) = Tf - 41.3

-0.256* Tf + 97.33*0.256 = Tf- 41.3

Tf(1+0.256) = 97.33*0.256+41.3

Tf = (97.33*0.256+41.3) / (1.256)

Tf = 52.7201 °C