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Eduardwww
2 months ago
13

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?
Physics
2 answers:
Softa [3K]2 months ago
7 0

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)
          ⇒ 32.5g (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

serg [3.5K]2 months ago
6 0

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

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