Question

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen boil away by the time the copper reaches 77.3 K? (The specific heat of copper is 0.092 0 cal/g · °C, and the latent heat of vaporization of nitrogen is 48.0 cal/g.)

Answer 1

1.2584kg of nitrogen boils.

Step-by-step explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

`U_1=U_2`

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so:
`U_1=U_1^C+U_1^(N_2)`

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

`U_2=U_2^C+U_(2,liq)^(N_2)+U_(2,vap)^(N_2)`

So, the overall energy balance is:

`U_1^C+U_1^(N_2)=U_2^C+U_(2,liq)^(N_2)+U_(2,vap)^(N_2)`

Reorganizing,

`U_1^C-U_2^C=U_(2,liq)^(N_2)+U_(2,vap)^(N_2)-U_1^(N_2)`

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

`m_C*Cp*(T_1^C-T_2^C)=m_(2,liq)^(N_2)u_(2,liq)^(N_2)+m_(2,vap)^(N_2)u_(2,vap)^(N_2)-m_1^(N_2)u_1^(N_2)`

Take in mind that, for the mass balance for nitrogen,
`m_1^(N_2)=m_(2,liq)^(N_2)+m_(2,vap)^(N_2)`,

So, let's replace
`m_1^(N_2)` in the energy balance:

`m_C*Cp*(T_1^C-T_2^C)=m_(2,liq)^(N_2)u_(2,liq)^(N_2)+m_(2,vap)^(N_2)u_(2,vap)^(N_2)-m_(2,liq)^(N_2)u_1^(N_2)-m_(2,vap)^(N_2)u_1^(N_2)`

So, as you can see, the term
`m_(2,liq)^(N_2)u_(2,liq)^(N_2)` disappear because
`u_(2,liq)^(N_2)=u_(1,liq)^(N_2)` (The specific energy in the liquid is the same because the temperature does not change).

`m_C*Cp*(T_1^C-T_2^C)=m_(2,vap)^(N_2)u_(2,vap)^(N_2)-m_(2,vap)^(N_2)u_1^(N_2)`

`m_C*Cp*(T_1^C-T_2^C)=m_(2,vap)^(N_2)(u_(2,vap)^(N_2)-u_1^(N_2))`

The difference
`(u_(2,vap)^(N_2)-u_1^(N_2))` is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

`m_(2,vap)^(N_2)=(m_C*Cp*(T_1^C-T_2^C))/((u_(2,vap)^(N_2)-u_1^(N_2)))`

`m_(2,vap)^(N_2)=(3kg*0.092(cal)/(gC) *(296.15K-77.3K))/(48.0(cal)/(g))\\m_(2,vap)^(N_2)=1.2584kg`