Calculate the density of liquid hydrogen if 4.224 kJ of heat is absorbed during the transition of 1 liter of hydrogen

| February 25, 2021 | education

Calculate the density of liquid hydrogen if 4.224 kJ of heat is absorbed during the transition of 1 liter of hydrogen at the boiling point to the gaseous state. The specific heat of vaporization of hydrogen is 59 kJ / kg.

To find out the required density of liquid hydrogen, we use the formula: Q = LH2 * m = LH2 * ρH2 * V, from where we express: ρH2 = Q / (LH2 * V).

Const: LH2 – beats. heat of vaporization of hydrogen (according to the condition LH2 = 59 kJ / kg = 59 * 10 ^ 3 kJ / kg).

Data: Q – absorbed heat (Q = 4.224 kJ = 4.224 * 10 ^ 3 J); V is the volume of evaporated hydrogen (V = 1 l; in the SI system V = 10 ^ -3 m3).

Let’s calculate: ρH2 = Q / (LH2 * V) = 4.224 * 10 ^ 3 / (59 * 10 ^ 3 * 10 ^ -3) ≈ 71.6 kg / m3.

Answer: The density of liquid hydrogen, according to the calculation, is 71.6 kg / m3.



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