A 60 g piece of aluminum at 20°C is cooled to -196°C by placing it in a large container of liquid nitrogen at that temperature. How much nitrogen is vaporized? (Assume that the specific heat of aluminum is constant and is equal to 0.90 kJ/kg·K and that the vaporized nitrogen's temperature does not change.)

Where C = Specific heat capacity of aluminum, M = mass of aluminum, t₂ = Final temperature, t₁ = initial temperature, c = latent heat of vaporization of nitrogen, m = mass of nitrogen.

make m the subject of the equation

m = CM(t₁-t₂)/c................ Equation 2

Given: C = 900 J/kg.K, M = 60 g = 0.06 kg, t₁ = 20 °C, t₂ = -196 °C

Constant: c = 199200 J/kg

Substitute these values into equation 2

m = 900×0.06×[20-(-196)]/199200

m = 900×0.06×216/199200

m = 0.0586 kg.

sebassandin sebassandin · Answer 2 · 2020-06-07T02:47:25+02:00

Answer:

[tex]m_{N_2}=58.6gN_2[/tex]

Explanation:

Hello,

In this case, for an average temperature of -176 °C, the vaporization enthalpy of liquid nitrogen is 199.2 J/g, thus, we first compute the heat lost by the aluminium by considering it cooled mass, specific heat and temperature change:

[tex]Q=mCp(T_2-T_1)=60g*0.90\frac{J}{g\°C}*(-196-20)\°C\\ \\Q=-11664J[/tex]

Next, heat lost by the aluminium is gained by the nitrogen:

[tex]-Q_{Al}=Q_{N_2}=11664J[/tex]

Therefore, the vaporized nitrogen is:

[tex]m_{N_2}=\frac{Q_{N_2}}{\Delta H_v}=\frac{11664J}{199.2J/g}\\\\m_{N_2}=58.6gN_2[/tex]

Best regards.