Question

A 26.5-g aluminum block is warmed to 65.2 ∘Cand plunged into an insulated beaker containing 55.2 g of water initially at 22.2 ∘C. The aluminum and the water are allowed to come to thermal equilibrium. (Cs,H2O=4.18 J/g⋅∘C, Cs,Al=0.903J/g⋅∘C)​

Assuming that no heat is lost, what is the final temperature of the water and aluminum?​

Answer 1

Initial temperature of water = 22.2 ⁰C

Initial temperature of aluminium = 65.2 ⁰C

let final temperature of water and aluminium = Tf

The heat absrobed by water = Heat loss by aluminium

Heat absorbed by water = Mass of water X specific heat of water X change in temperature = Qw

Qw = 55.2 X 4.18 X (22.2-Tf)

Heat loss by metal = Mass of aluminium X specific heat of aluminium x change in temperature = QAl

QAl = 26.5 X 0.903 X (65.2 -Tf)

equating the two heat

26.5 X 0.903 X (65.2 -Tf) = 55.2 X 4.18 X (22.2-Tf)

23.93 X (65.2-Tf) = 230.74 X (Tf - 22.2)

1560.24 - 23.93T = -5122.34 + 230.74 Tf

6682.58 = 254.67x Tf

Tf = 26.24 ⁰C