If 420 J of heat is added to a 32.0 g sample of liquid methane under...

If 420 J of heat is added to a 32.0 g sample of liquid methane under 1 atm of pressure at a temperature of -170 degrees C, what is the final phase of methane when the system equilibrates? Assume no heat lose. Normal BP = -161.5 degrees C

Specific heat (liq.) = 3.48 J/gC

specific heat (gas) = 2.22 J/gC

Heat of Vaporization = 8.20 kJ/mol

Solutions

Expert Solution

heat supplied will be used to heat liqudi methane from -170 deg.c to -161.5, the boiling point and this heat is in the form of sensible heat, sensible heat =mass of liquid methane* specific heat* temperatrue difference= 30*3.48*(-161.5-(-170)= 887.4 joules

however, heat supplied is only 420 joules while 887.4 is required to take the liquid to boiling point. So the liquid methane with the supplied heat does not reach even the boiling point and hence it suggests the liquid remains as liquid only and its temperature (T) can be calculated as

30*3.48*(T+170)= 420

T+170= 420/(30*3.48), T= -165.977 deg.c

since there are no heat losses when the system equilibrates, methane remains as liquid at -165.977 deg.c

queen_honey_blossom answered 1 year ago

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