Question

how much heat is needed to convert 10.0 grams of CH3OH(l) at -0.4 C to CH3OH(g) at 64.6 C?

Answer 1

Given that;

Mass ; m = 10.0 g CH3OH

T 1= - 4.0 C

T2= 64.6 C

The heat capacity of CH3OH = 0.328 J/g⋅∘C

Heat of vaporization of CH3OH =35.21 KJ/mole

Its boiling point is 64.6 degrees Celsius.

This is actually two different steps to to warm 10.0g of CH3OH initially at -4.0 ∘C, to VAPOR at 64.6 ∘C which are following:

1) vaporizing the CH3OH to form gas CH3OH
Heat of vaporization of CH3OH =35.21 KJ/mole

Therefore calculate the number of moles as follows:

Number of moles = amount in g / molar mass

= 10.0 g / 32.0419 g CH3OH/mol

= 0.312 moles
q1 = heat of vaporization * number of moles

= 35.21 KJ/mole * 0.312 moles

= 10.99 KJ

2) heating the Vpaor up to 64.6 C

q5 = m*c*dT

here m = mass of CH3OH

c = specific heat of CH3OH

dT = temperature change

Here dT= 64.6- (-4.0)= 68.6 C

Then;

q2 = 10 g * 0.328 J/g C * 68.6 C

= 225.008 J

= 0.225 kJ

Now, add up the heat required to get the final answer:
Q = q1+ q2

Q= 10.99 KJ + 0.225 KJ

= 11.215 KJ