A 15.7-g aluminum block is warmed to 53.2 °c and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 °c. the aluminum and the water are allowed to come to thermal equilibrium. assuming that no heat is lost, what is the final temperature of the water and the aluminum?

Answer: 27.2 °C

Explanation:

1) Physical principles:

a) First law of thermodynamics: energy is conserved

b) Insulated system ⇒ no heat is lost ⇒ Q gained by water = Q lost by aluminiun

c) Thermal equilibrium: final temperature of water = final temperature of aluminum.

2) Formula:

Q (gained or released) = m*Cs*ΔT, where m is the mass of the substance, Cs is the specific heat of the substance, and ΔT is the change in temperature

3) dа ta:

mass aluminum, Ma = 15.7g Ti, a = 53.2 °C mass water, Mw = 32.5 g Ti, w = 24.5°C

4) Information from tables (internet)

Specific heat liquid water: Cs = 1 cal/g°C Specific heat aluminum: Cs = 0.215 cal/g°C

5) Solution:

Q water = Q aluminun

Qwater = Mw*Cs*ΔT = 32.5g (1 cal/g°C) (Tf - 24.5°C)

Q aluminum = Ma*Cs*ΔT = 15.7g (0.215cal/g°C) (53.2°C - Tf) ⇒ 32.5g (1 cal/g°C) (Tf - 24.5°C) = 15.7g (0.215cal/g°C) (53.2°C - Tf)

⇒ 32.5Tf - 796.25 = 179.5766 - 3.3755Tf

⇒ 35.8755Tf = 975.8266 ⇒ Tf = 27.2°C ← answer

The equilibrium temperature of aluminium and water is 33.2°C

We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K

Now we can calculate the equilibrium temperature

(mc∆T) _aluminium = (mc∆T) _water

15.7*0.9 * (53.2-T) = 32.5*1 * (T-24.5)

T=33.2°C