Answer to Question #116527 in Mechanics | Relativity for Augustine

Question

A copper calorimeter of mass 150g was half-filled with water of mass 300g and temperature 0°C. 5g of ice at 0°C was added to the content, later some quantity of steam was passed into the mixture and the temperature rose by 20°C. Calculate the quantity of steam added? (Specific latent heat of vaporization of steam = 2260000j/kg, specific heat of copper = 400J/kg/K)

Answer 1

Let us write the equation of thermodynamic equilibrium:

$c_cm_c\Delta T + c_w(m_w+m_i)\Delta T + m_i\lambda_i = m_sL_s + m_sc_w\Delta T_2$ .

Here $L_s = 2260000\,J/kg, \; c_c = 400\,J/kg/K, \;\; c_w = 4200\, J/kg/K, \;\; c_i = 2100\; J/kg/K, \;\; \lambda_i = 330000\, J/kg.$

Next, we obtain the expression for the mass of steam

$m_s = \dfrac{c_cm_c\Delta T + c_w(m_w+m_i)\Delta T + m_i\lambda_i }{L_s + c_w\Delta T_2} = \\ = \dfrac{400\cdot0.150\cdot20 + 4200\cdot(0.3+0.005)\cdot20 + 0.005\cdot330000}{2260000 + 4200\cdot80} \approx 0.011\,\mathrm{kg}.$