Ahouse has well-insulated walls. it contains a volume of 135 m3 of air at 275 k. (a) consider heating it at constant pressure. calculate the energy required to increase the temperature of this diatomic ideal gas by 1.8°c. kj (b) if this energy could be used to lift an object of mass m through a height of 2.4 m, what is the value of m? if this energy could be used to lift an object of mass m through a height of 2.4 m, what is the value of m? kif this energy could be used to lift an object of mass m through a height of 2.4 m, what is the value of m? kg

Energy required = 313.3562 kJ

Mass that can be lifted = 13309.38 Kg

Explanation:

The general equation to get the heat transferred at constant pressure is  

Q = m* cp* delta T   (equation 1)

Where:

Q = heat transfer (Joules, J)

m= mass of the substance (Kg)

Cp = specific heat (J/kg K)  

Delta T = change in temperature is (Tfinal -Tinitial)

The temperature in the house is 275K. There is not given value for the pressure but since there is no information that tells you that the house is being in a condition of low or high pressure, we can use standard conditions, which are standardized conditions for pressure and temperature that allows comparisons between data. For pressure the value is 101.325 kPa.

The air is a mix of mainly two gases N2 and O2 with a very small among of other gases that don’t need to be considering in the calculations. We can consider this mix as an ideal gas which mean we don’t consider the effect of the intermolecular forces between the molecules.  The molecular weight of the air is 28,966 g/mol.

We can get the mass of the air in the room multiplying the volume given by the density (ρ) of the air at the condition of 275K and 101.325 kPa.  

The air’s density for this condition is ρ= 1.285 kg/m^3. You can get this value from thermodynamic tables for air.  

Air’s mass in the room = 135m^3* 1.285 kg/m^3 = 173.475 Kg

Cp is a property that change with temperature. For air we can use the next equation to get the value for this property at different temperatures.

Cp/R = α + βT + γT^2 + δT^3 +  εT^4   (equation 2)

Where:

R is the gas constant (8.314 J/mol K)

α, β,γ,δ and ε are constant with the values:

α=3.653

β=-1.337*10^(-3)

γ=3.294*10^(-6)

δ=-1.913*10^(-9)

ε=0.2763*10^(-12)

T= temperature in K

Since there is a change in the temperature in this process, we are going to calculate the Cp for both temperature (275K and 276.8 K (delta in °C is the same in K)) and then we use the average to calculate the energy.

Using the equation 2 we get the next values for Cp/R  

Cp/R (T= 275K) = 3.4962

Cp/R (T= 276.8K) = 3.4963

Cp (T=275K) = 8.314 J/mol K* 3.4962 = 29.0676 J/mol K

Cp (T=276.8K) = 8.314 J/mol K* 3.4963 = 29.0686 J/mol K

This value is molar. To get the Cp in mass units we divide by the molecular weight of air (28,966 g/mol) and multiply by 1000 to get the value in Kg  

Cp (T=275K) = (29.0676 J/mol K) * (1 mol /28.966 g)* (1000 g /1 Kg) = 1003.509 J/kg K

Cp (T=276.8K) = (29.0686 J/mol K) * (1 mol /28.966 g)* (1000g /1 Kg) = 1003.543 J/kg K

Average = [Cp (T=275K) + Cp (T=276.8K)] /2

average =  1003.526 J/kg K  

Energy transferred  

Q = m* cp* delta T    (equation 1)

With  

m = 173.475 Kg

Cp = 1003.526 J/kg K  

Delta T =1.8 K. delta T in °C is the same in K  

Q = 173.475 Kg * 1003.526 J/kg K * 1.8 K = 313356.219 J or 313.3562 kJ

Since work is closely related to energy, we can use the next equation to calculate the mass that can be lift with this energy

W or E (J) = F*s  

Where :

W= work in joules

E = energy  in joules

F is the mass in newtons – 1 kg is equal to 9.81 newtons  

S is the height – it is 2.4 meters in this case  

So  

F= E/s = 313356.219 J / 2.4 m = 130565.0916 newtons  

Mass in kg = 130565.0916 newtons * 1 Kg /9.81 newtons = 13309.38 Kg  

hi

explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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