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TheProfessor4zzzFM
You just need the energy of turning of water in to steam. Water is a little funny because once water gets to 100 degrees C, it stops changing in temperature whilst it all turns to steam. The amount of energy required to turn 1 gram of water into steam is 2257 Joules. We call this the latent heat of vaporisation. So to turn 74.6 grams of water to steam, you would need:

2257J/gram x 74.6g = 168,372.2 J = 168.4 kJ

(latent heat of vaporisation: 2257 J/gram)


okpalawalter8

We have that the amount of heat (Q) is mathematically given as

Q=169 kJ

Amount of heat

Question Parameters:

the amount of heat (in kj) required to convert 74.6 g of water to steam at 100°c

Generally the equation for the number of moles of water  is mathematically given as

number of moles of water= g of water / molar mass of water

N = 74.6g of H2O*1 mol of H2O / 18.02 g of H2O

N=4.14

Then multiply the number of moles by the molar heat of vaporization of water.

Q=molar heat *moles

Q = (40.79) * (4.14)

Q=169 kJ

For more information on Heat visit

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