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 °

Answer: 27.2 °C

Explanation:

1) Physical principles:

a) The first law of thermodynamics states that energy is conserved.

b) An insulated system means no heat loss, hence the heat gained by water equals the heat lost by aluminum.

c) In thermal equilibrium, both the final temperatures for water and aluminum are the same.

2) Formula:

Q (gained or released) = m×Cs×ΔT, where m represents the substance's mass, Cs its specific heat, and ΔT the temperature change.

3) Data:

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

4) Information from tables (internet)

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

5) Solution:

• Q water = Q aluminum

• 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)

    ⇒ 32.5Tf - 796.25 = 179.5766 - 3.3755Tf

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

The equilibrium temperature of aluminum and water is 33.2°C

The specific heat for aluminum is 0.9 J/gm-K, while for water it is 1 J/gm-K.

Let's determine the equilibrium temperature:

(mc∆T)_aluminum=(mc∆T)_water

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

T=33.2°C