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Today, 04:43

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?

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Answers (2)
  1. Today, 07:40
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    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
  2. Today, 08:37
    0
    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
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