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A 1.00-kg iron horseshoe is taken from a forge at 900∘C and dropped into 4.00 kg of water at 10.0∘C. Assuming that no energy is lost by heat to the surroundings, determine the total entropy change of the horseshoe-plus-water system.

Respuesta :

Answer:

[tex]dS=1379.3295\ J.K^{-1}[/tex]

Explanation:

Given:

  • mass of iron, [tex]m_s=1\ kg[/tex]
  • initial temperature of iron, [tex]T_i_s=900^{\circ}C[/tex]
  • mass of water, [tex]m_w=4\ kg[/tex]
  • initial temperature of water, [tex]T_i_w=10^{\circ}C[/tex]

We have,

  • Specific heat of iron, [tex]c_s=448\ J.kg^{-1}.K^{-1}[/tex]
  • Specific heat of water, [tex]c_w=1486\ J.kg^{-1}.K^{-1}[/tex]

Now the final temperature of the system assuming that no heat is lost to the surrounding:

[tex]m_s.c_s.(T_i_s-T_f)=m_w.c_w.(T_f-T_i_w)[/tex]

[tex]1\times 448\times (900-T_f)=4\times 4186\times (T_f-10)[/tex]

[tex]403200-448\times T_f=16744\times T_f-167440[/tex]

[tex]T_f=14.467\ ^{\circ}C[/tex]

Now the change in heat energy:

[tex]dQ=m.c_s.\Delta T[/tex]

[tex]dQ=1\times 448\times (900-14.467)[/tex]

[tex]dQ=396718.6254\ J[/tex]

Hence the change in the entropy of the system:

[tex]dS=\frac{dQ}{T_f}[/tex]

[tex]dS=\frac{396718.6254}{14.467+273.15}[/tex]

[tex]dS=1379.3295\ J.K^{-1}[/tex]

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