Answer in Molecular Physics | Thermodynamics for Rahma #144947
February 21st, 2023
(a)
“varphi_1=frac{rho_1}{rho_0}cdot 100%=50%.”
From the table of dependence of saturated steam density on temperature “t=20u00b0C, rho_0=17.3 frac{g}{m^3}.”
“rho_1=frac{varphi_1rho_0}{100}=frac{50cdot17.3}{100}=8.65 frac{g}{m^3}.”
“m_1=rho_1V=8.65cdot20=173 enspace g.”
(b)
“m_2=rho_2V=frac{varphi_2rho_0}{100}V=frac{30}{100}cdot17.3cdot20=103.8enspace g.”
“Delta m=m_1-m_2=173-103.8=69.2enspace g.”
It can be putted into a cup.
