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Triss [41]
2 years ago
11

13. Mercury has one of the lowest heat capacities. If 257J of heat are added to 5500g of mercury,the mercury's temp will increas

e by 5.5 degrees Celsius. What is the specific heat capacity of mercury?
Chemistry
1 answer:
NikAS [45]2 years ago
3 0

Given:

257J of heat

5500g of mercury

increase by 5.5 degrees Celsius

Required:

Specific heat of mercury

Solution:

H = mCpT

257J = (5500g of mercury) Cp (5.5 degrees Celsius)

Cp = 8.5 x 10^-3 Joules per gram per degree Celsius

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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
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Answer:

1.2584kg of nitrogen boils.

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}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}

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

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