Question

How much energy is required to heat 36.0 g H2O from a liquid at 55.0°C to a gas at 150.0°C? The following physical data may be useful. Molar Mass(H2O) = 18.0 g/mol

ΔH_vap = 40.7 kJ/mol

            C_s;liquid = 4.18 J/g ^oC

            C_s;gas = 2.01 J/g^oC

            T_b(H₂O) = 100.0 ^oC

Answer 1

For calculation of energy required to heat 36.0 gms of water from a liquid at 55.0 ⁰c to a gas at 150.0 ⁰c,

first we will heat the water from 55⁰c to its boiling point 100 ⁰c,

we know that the formula,

q = m*c*(T2-T1),

by putting the values to the formula,

q = 36.0*4.18 J/g C */(100-55), ( where m= mass of water in gms, C LIQUID= 4.18 J/g C )

q = 36.0*4.18*45,

q = 6771.6 J , ( 6771.6/1000=6.7716 KJ)

NOW we have to vaporise the water 10 100 ⁰ c,

q = 36/18.0*40.7 kj/mol (where moles of water * delta Hvap),

q = 2*40.7=80.4 KJ,

NOW WE HAVE TO HEAT THE WATER (GAS) TO 155 ⁰ C FROM 100 ⁰ C,

q = m*c,

q = 36.0 *2.01 J/g C(T2-T1),

q = 36*2.01J/g C(155-100),

q = 36.0*2.01*55,

q = 3979.8 J (3979.8/1000=3.9798 KJ)

now we will add up all the energies by converting it to the KJ,

Liquid (6.7716 KJ)+ VAPOURISATION (80.4 KJ)+ GAS ( 3.9798 KJ) = 91.1514 KJ,

The total energy required = 91.1514 KJ