Question

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.

Answer 1

`dS=1379.3295\ J.K^(-1)`

Step-by-step explanation:

Given:

• mass of iron,
  `m_s=1\ kg`

• initial temperature of iron,
  `T_i_s=900^(\circ)C`

• mass of water,
  `m_w=4\ kg`

• initial temperature of water,
  `T_i_w=10^(\circ)C`

We have,

• Specific heat of iron,
  `c_s=448\ J.kg^(-1).K^(-1)`

• Specific heat of water,
  `c_w=1486\ J.kg^(-1).K^(-1)`

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

`m_s.c_s.(T_i_s-T_f)=m_w.c_w.(T_f-T_i_w)`

`1* 448* (900-T_f)=4* 4186* (T_f-10)`

`403200-448* T_f=16744* T_f-167440`

`T_f=14.467\ ^(\circ)C`

Now the change in heat energy:

`dQ=m.c_s.\Delta T`

`dQ=1* 448* (900-14.467)`

`dQ=396718.6254\ J`

Hence the change in the entropy of the system:

`dS=(dQ)/(T_f)`

`dS=(396718.6254)/(14.467+273.15)`

`dS=1379.3295\ J.K^(-1)`