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Question: How many joules of heat must be added to a \( 2kg \) sample of methanol for it to boil?...

How many joules of heat must be added to a 2kg2kg sample of methanol for it to boil?

Explanation

Solution

Heat transfer is an important phenomenon, the energy needed to break interatomic attraction during boiling or melting of the system is known as latent heat. The temperature remains constant during this change. Specific heat capacity is defined as the amount of heat which is required to increase the temperature of one kilogram of a substance by one kelvin.

Complete answer:
For methanol to boil we need to supply heat till its boiling point 64.7oC{64.7^o}C
The total heat input should be
Q=m.Cp.dT+HvQ = m.{C_p}.dT + \partial {H_v}
Where, Q=Q = heat transfer, m=m = mass of substance, dT=dT = change in temperature, Hv=\partial {H_v} = latent heat of evaporation.
m=2kg Mw=32gmol molmethanol=20032=62.5mol  m = 2kg \\\ {M_w} = 32\dfrac{g}{{mol}} \\\ mo{l_{methanol}} = \dfrac{{200}}{{32}} = 62.5mol \\\
We know Cp=79.5Jmol.K{C_p} = 79.5\dfrac{J}{{mol.K}} ( Cp={C_p} = specific heat at constant pressure.)
And we know boiling point of methanol, Tboil=64.7oC{T_{boil}} = {64.7^o}C and Hv=38.278kJmol\partial {H_v} = 38.278\dfrac{{kJ}}{{mol}}
Sensible heat, Q=62.5×79.5×(64.720) 222.1kJ  Q = 62.5 \times 79.5 \times (64.7 - 20) \\\ 222.1kJ \\\ ( considering initial temperature , To=20oC{T_o} = {20^o}C )
Latent heat, 38.278×62.5=2392.37kJ38.278 \times 62.5 = 2392.37kJ
Total heat == sensible heat ++ latent heat
Total heat
=222.1+2392.37 =2614.47kJ  = 222.1 + 2392.37 \\\ = 2614.47kJ \\\

Note:
Sensible heat and latent heat are types of energies absorbed or released in the atmosphere. Sensible heat is related to changes in temperature of a gas or objects with no change in phase. Latent heat is change in phase between liquids, solids and gases at constant temperature.