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Here's a focus problem for you to work on. You have 1 kg of liquid water, saturated at 50 C, sitting under the barrier in the large, rigid, adiabatic container, with vacuum above the barrier. You remove the barrier and allow the system to re-equilibrate. For the final equilibrium temperature of the resulting liquid water and water vapor mixgture to be 45 C,
1. What is the final mass of water vapor in the head space?
2. What is the total volume of the rigid container?
For using the steam tables to solve this, the applicable equations would be:
$$m_L+m_V=m_0$$
$$m_Lu_L+m_Vu_V=m_0u_{L0}$$
$$m_Lv_L+m_Vv_V=V$$
where ##m_0## is the original mass of liquid water (1 kg), ##u_{L0}## is the original internal energy per kg of the original liquid water at 50 C, V is the total volume of the rigid container, the lower case v's are specific volume (volume per kg) and where lower case symbols without subscript 0's refer to the final state..
Show that these equations can be reduced to:
$$u_L+x(u_V-u_L)=u_{L0}$$
$$v_L+x(v_V-v_L)=\frac{V}{m_0}$$
where x is the equilibrium mass fraction of vapor at the new temperature of 45C.
1. What is the final mass of water vapor in the head space?
2. What is the total volume of the rigid container?
For using the steam tables to solve this, the applicable equations would be:
$$m_L+m_V=m_0$$
$$m_Lu_L+m_Vu_V=m_0u_{L0}$$
$$m_Lv_L+m_Vv_V=V$$
where ##m_0## is the original mass of liquid water (1 kg), ##u_{L0}## is the original internal energy per kg of the original liquid water at 50 C, V is the total volume of the rigid container, the lower case v's are specific volume (volume per kg) and where lower case symbols without subscript 0's refer to the final state..
Show that these equations can be reduced to:
$$u_L+x(u_V-u_L)=u_{L0}$$
$$v_L+x(v_V-v_L)=\frac{V}{m_0}$$
where x is the equilibrium mass fraction of vapor at the new temperature of 45C.