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]
