Question #116527
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)
1
Expert's answer
2020-05-20T09:26:56-0400

Let us write the equation of thermodynamic equilibrium:

ccmcΔT+cw(mw+mi)ΔT+miλi=msLs+mscwΔT2c_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 Ls=2260000J/kg,  cc=400J/kg/K,    cw=4200J/kg/K,    ci=2100  J/kg/K,    λi=330000J/kg.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

ms=ccmcΔT+cw(mw+mi)ΔT+miλiLs+cwΔT2==4000.15020+4200(0.3+0.005)20+0.0053300002260000+4200800.011kg.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}.


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Comments

Faithnix
22.08.23, 00:08

Very good thanks it really helped

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