Question

calculate the change in thermal energy in kJ for a 637 g sample of acetone (molar mass = 58.08 g/mol) to change from -10.3 Celsius to 108 Celsius.

given information:

enthalpy of fusion: 5.73 kJ/mol

enthalpy of vaporization: 31.3 kJ/mol

melting point: -94.7 C

boiling point: 56.1 C

specific heat capacity for solid: 1.65 J/g Celsius

specific heat capacity for liquid: 2.16 J/g C

specific heat capacity for gas: 1.29 J/g C

thank you!!

Answer 1

This is advance specific heat problem. there are two heat formula needed for this problem .But before we jump into the calculation let we understand first what is requirements is and why we needed two formula.

. Any substance has generally three state solid ,liquid and gaseous and it is temperature dependent. Different temperature different state occurs. We also called this state as phase. Now when specific temperature applied , solid form become liquid and this particular point or value of temperature is called melting point of that substance .simllar liquid to gaseous phase achieved by supplying temperature of boiling point of that particular substance.

In melting point time and boiling point time the solid -liqiud and liquid - gaseous phases are occurs simultaneously. This is called laten heat of fusion ans laten heat of vaporization . Where the heat formula different with constant temperature.the following diagram improve your understanding

heat of vaporization gas) -> boiling point (T=constant) Temperature → heat of (Liquid) fusion ..) melting point (temperature (1) constan) is (solid) Energy supplied

The two formulas are

   q= m × c_p × ∆T ----(1)

q = m × L. ------- (2)

where, q= heat of the substance ; m = mass of the substance ; c_p= specific heat of the substance ; L = laten heat of the substance .

.

the next diagram for acetone ,you can understand when this formula needed

94.7°c melting point 10827 56.1 °C boiling point 个 56.121 a design (gas) i= m xq xat) boiling point q = mL temp (Liquid) -10.30 9 = mx cpxat -94.7°C Melting point Time →

acetone already in liquid form,no need of heat of fusion.

Let's do it .

. ( Calculate the heat for every phases then sumup all for final answer.)

(A) For liquid state : AT = 977m x CP X AT (56.1 - 6-10.3)) c = 66.4 °c 2.16 519°C Cp (liquid) m= 637 g. q 6379X 2.16 J/go¢ X 66.4°C = 91361.088 J✓ (B) For Laten heat of vaporization: m=6379 6378 :. mole = 58.02 g/m.c! 9= mxL L = 31.3 kJ/mol = 31300 T/mol 10.96 mol

92 = 10.96 mol x 31300 1/myl : 343286.847 5 (* We convent 6379 ac tone into mole using molar mass multiply by 1000.) and KJ to I (C) For gaseous state 9= mx Cpx AT me 6379 Cp (gas) = 1.29 ulge (108 - 56.1)°C = 5190 OT - 11 9 = 6378 x 1-29 J/gt * 51.9% = 42647.787 T
the mole number is too accurate (7 decimal place 10.9676309 mol)so heat q₂ was determine with multiply with L value)

Total 9, +92 +93 = 91361.088 J + 343286:847 J 42647: TST T 477295.7225

4777295.722 J or 477.295 kJis answer.

Hope you understand.thank you!!