Question

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.)

Answer 1

`m_(N_2)=58.6gN_2`

Step-by-step 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:

`Q=mCp(T_2-T_1)=60g*0.90(J)/(g\°C)*(-196-20)\°C\\ \\Q=-11664J`

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

`-Q_(Al)=Q_(N_2)=11664J`

Therefore, the vaporized nitrogen is:

`m_(N_2)=(Q_(N_2))/(\Delta H_v)=(11664J)/(199.2J/g)\\\\m_(N_2)=58.6gN_2`

Best regards.

Answer 2

0.0586 kg

Step-by-step explanation:

From the question,

Heat lost by the aluminium = heat gain by nitrogen.

CM(t₁-t₂) = cm................... Equation 1

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.