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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.

Learn more about Specific heat of water :

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alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.