A rigid cylinder contains 9.0 g of nitrogen at 20 C. What is the minimum amount

Question

A rigid cylinder contains 9.0 g of nitrogen at 20 C. What is the minimum amount of heat energy that must be removed to liquify the nitrogen? Express your answer with the appropriate units.

T = -20C, (The initial Temperature of ice cube)

C ice = 2090J/kg K , (The specific heat of ice)

C water = 4186J/kg K, (The specific hear of water)

C steam = 2009 J/kg K, (The specific heat of steam)

Lf = 3.33X10^5 J/kg, (The latent heat of fusion of ice)

Lv = 2.3 X 10^5 J/kg (The latent heat of vaporization of steam)

Explanation / Answer

let;

H = Heat Energy ;Cv = specific Heat; m = mass (in Kg); T = change in temperature (in Celsius) ;Heat of Condensation (Equal to Vaporisation) ;

H = Hc * m

Hc= Heat of condensation ;m= mass (in moles)

formula;

Heat Energy H = Cv * m * T

Heat Capacity of nitrogen =0.743kJ/kg*K.

Nitrogen is a liquid at -196 C,

so T = -196 - 20 = -216

H = Heat Capacity * mass * T

H = 0.743 kJ/kg*K * 0.009 kg * 216 = 1.444 kJ

H= Heat of condensation * mass (in moles)

H= 5.56 kJ/mol * 0.32 moles = 1.7792 kJ

1.444 kJ + 1.7792 kJ = 3.223 kJ

ANSWER;

3.223 kJ of heat energy is required to liquify the nitrogen

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