Respuesta :
Answer:
Q1 = C * m * dT
Q2 = Qm * m
Qtotal = Q1 + Q2
Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.
C - is calorific capacity of oxygen -- better look at tables, it is a constant value
m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.
dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)
Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)
Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions
If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0
I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.
Data from tables
Tm = 54.36 K
C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.
Qm = 0.444 kJ/mol
Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol
So you have 44.33/32 = 1.385 moles of oxygen
a) if oxygen is already at melting temperature: you only have to melt it
Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J
b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.
Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J
Q2 = Qm * m = 615 J
Qtotal = 177.442 + 615 = 792.44 J
Explanation:
Answer:
In this case, 1 mole of liquid oxygen requires 3.41 kJ of heat in order to perform that liquid → solid. your sample from grams to moles by using oxygen's molar mass.
