Answer to Question #270516 in Molecular Physics | Thermodynamics for Tipu

Question

a lump of steel of mass 8 kg at 1000 k is dropped in 80 kg of oil at 300 k. calculate the entropy change of steel the oil and the universe. take specific heats of steel and oil as 0.5 kj/kg K and 3.5 Kj /kg K respectively

Answer 1

$m_s = 8 \;kg \\ m_o = 80 \; kg \\ T_1 = 1000 \;K \\ T_2 = 300 \;K \\ c_{ps} = 0.5 \; kJ/kg \cdot K \\ c_{po} = 3.5 \; kJ /kg \cdot K$

Heat lost (steel) = Heat gain (Oil)

$m_s c_{ps}(T_1 -T) = m_o c_{po} (T- T_2) \\ 8 \times 0.5 \times (1000 -T) = 80 \times 3.5 (T- 300) \\ 4000 -4T = 280T -84000 \\ 88000 = 284T \\ T = 309.8 \;K$

Entropy change

System: steel $= m_s c_{ps} ln(\frac{T}{T_1})$

$= 8 \times 0.5 \times ln(\frac{309.8}{1000}) = -4.68 \; kJ/K$

Surrounding: Oil $= m_o c_{po} ln(\frac{T}{T_2})$

$= 80 \times 3.5 \times ln(\frac{309.8}{300}) = 9.00 \; kJ/K$

Universe = System – Surrounding

$= -4.68 + 9.00 = 4.32 \;kJ/K$