Answer to Question #210119 in Molecular Physics | Thermodynamics for Tonnoy

Question

A piece of steel of mass 35.0kg is taken out of a furnace at 820degree C and dropped into oil held in a steel container of mass 20.0kg. The oil and container are at 16.0degree C. For safety reasons, the final equilibrium temperature of the steel, oil and container must not exceed 40.0degree C. (Specific heat capacities are: steel 450J/kg/K and oil 2700J/kg/K)

what is the minimum mass of the oil?

Answer 1

$c_sm_s(t_2-t)=(c_sm_s'+c_om_o)(t-t_1),\implies$

$m_o=\frac{c_sm_s(t_2-t)}{c_o(t-t_1)}-\frac{c_sm_s'}{c_o}=186.25~kg.$