PLEASE HELP ASAP WILL GIVE BRAINLIEST (dont give chegg link pls)

How much heat is required to convert 150 g of liquid H2O at 43.5 °C into steam at 130 °C?

Specific heat of liquid water is 4.18 J/g oC.
Specific heat of water vapor is 1.84 J/g oC.
ΔHvap = 40.67 kJ/mol

The amount of heat required to convert H₂O to steam is :  382.62 kJ

Given data :

Mass of liquid water  ( m ) = 150 g

Temperature of liquid water = 43.5°C

Temperature of steam = 130°C

The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )

where ;

q1 = heat required to change Temperature of water from 43.5°C to 100°C .  q2 = heat required to change liquid water at 100°C to steam at 100°C

q3 = heat required to change temperature of steam at 100°C to 130°C

M* S*ΔT

= 150 * 4.18 * ( 100 - 43.5 )

= 35425.5 J

moles * ΔHvap

= (150 / 18 )* 40.67 * 1000

=  338916.67 J

M * S * ΔT

= 150 * 1.84 * ( 130 -100 )

= 8280 J

Back to equation ( 1 )

Amount of heat required = 35425.5  + 338916.67 + 8280 = 382622.17 J

                                                                                               ≈ 382.62 kJ

Hence we can conclude that The amount of heat required to convert H₂O to steam is :  382.62 kJ.

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