How much heat is required to convert 422 g of liquid h2o at 23.5 °c into steam at 150 °c?

The heat that is required to convert 422 g of liquid H₂O at 23.5 °c into steam at 150 °c can be calculated as follows:

The heat capacity of water = 4184 J K⁻¹ C⁻¹

Therefore, heat required to warm the water from

23.5 °C to 100.0 °C.

                         =

Here,

m=0.422 kg

c=4184 J K⁻¹ C⁻¹

=100-23.5

   so, heat required to warm the water from  23.5 °C to 100.0 °C

                 = 0.422 × 4184 × (100-23.5)

                        = 135072.072 J

Latent heat of vaporization of water is 2260 kJ/kg

Thus the heat will be 0.422 × 2260000 = 953720 J

Heat required to raise steam from 100 to 150 

   2000 × 0.422 ×50 = 42200 J

Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

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