Answer to Question #216285 in Molecular Physics | Thermodynamics for Ope

Question

a lump of steel weighing 30kg at a temp of 427 degree celcius is dropped into 150kg of oil at 27 degree celsius. The specific heat capacity of the steel and the oil are 0.47 KJ/KgK and 2.5 KJ/KgK

Answer 1

Gives

$$m_s=30kg,T=700K,c_s=0.47KJ/kg-K$$

$$m_o=150kg,c_o=2.5KJ/kg-K,T=300K$$

${(ms∆T)}_{s}={(ms∆T)}_{o}$

$$30\times0.47\times(700-T)=150\times(T-300)\times2.5$$

$14.1(700-T)=375(T-300)$

$T=315K$

entropy steel

$$(∆s)_s=\smallint_1^2\frac{dQ}{T}=\int_1^2\frac{mc_sdT}{T}=mc_sln\frac{T_2}{T_1}$$

$$(∆s)_s=30\times0.47\times ln\frac{315}{700}=-11.25kJ/K$$

Similarly

$$(∆s)_o=150\times2.5\times ln\frac{315}{300}=18.29kJ/K$$

$$∆s_{univese}=∆s_s+∆s_o=-11.25+18.29=7.03KJ/K$$