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
Answer:
The final temperature of the water–horseshoe system is 30⁰C
Explanation:
Given;
mass of iron, [tex]M_f_e[/tex] = 1.50-kg
initial temperature of the iron horseshoe, [tex]T_f_e[/tex] = 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
[tex]M_f_eC_f_e(550 -T) = M_wC_w(T-24)[/tex]
where;
[tex]C_f_e[/tex] is the specific heat capacity of iron = 449 J/kg.°C
[tex]C_w[/tex] is the specific heat capacity of water = 4200 J/kg°C
[tex]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[/tex]
Therefore, the final temperature of the water–horseshoe system is 30⁰C
