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?

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

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