A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 14.0 kg of water at 24.0°C. What is the final temperature of

Question

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 14.0 kg of water at 24.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

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Diễm Thu 5 years 2021-08-04T13:31:13+00:00 2 Answers 71 views 0

Answers ( )

  1. thienthanh
    0
    2021-08-04T13:32:33+00:00
    Reply

    Answer:

    29.97 °C

    Explanation:

    From the question,

    Heat lost by the iron horseshoe = heat gained by water.

    CM(t₁-t₃) = cm(t₃-t₂)………………. Equation 1

    Where C = specific heat capacity of the iron horseshoe, M = mass of the iron horse shoe, c = specific heat capacity of water, m = mass of water, t₁ = Initial temperature of the iron horseshoe, t₂ = initial temperature of water, t₃ = final Temperature of the mixture.

    make t₃ the subject of the equation,

    t₃ = (CMt₁+cmt₂)/(CM+cm)…………………. Equation 2

    Given: M = 1.5 kg, m = 14 kg, t₁ = 550 °C, t₂ = 24 °C

    Constant: C = 450 J/kg.K, c = 4200 J/kg.K

    Substitute into equation 2

    t₃ =[(450×1.5×550)+(4200×14×24)]/(450×1.5+4200×14)

    t₃ = (371250+1411200)/(675+58800)

    t₃ = 1782450/59475

    t₃ = 29.97 °C

    Hence the final temperature of water = 29.97 °C

  2. RuslanHeatt
    0
    2021-08-04T13:33:06+00:00
    Reply

    Answer:

    The final temperature of the water–horseshoe system is 30⁰C

    Explanation:

    Given;

    mass of iron, M_f_e = 1.50-kg

    initial temperature of the iron horseshoe, T_f_e = 550°C

    mass of water, Mw =  14.0 kg

    initial temperature of water,Tw = 24°C

    Let the final temperature of the water–horseshoe system = T

    From the principles of conservation of heat;

    Heat lost by a hot body = Heat gained by a cold body

    Heat lost by hot iron horseshoe = heat gained by water

    M_f_eC_f_e(550 -T) = M_wC_w(T-24)

    where;

    C_f_e is the specific heat capacity of iron = 449 J/kg.°C

    C_w is the specific heat capacity of water = 4200 J/kg°C

    1.5*449(550 -T) = 14*4200(T-24)\\\\370425 - 673.5T = 58800T - 1411200\\\\370425 + 1411200 = 58800T + 673.5T\\\\1781625 = 59473.5T\\\\T = \frac{1781625}{59473.5} = 29.96 \ ^oC = 30\ ^oC

    Therefore, the final temperature of the water–horseshoe system is 30⁰C

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