Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are δhs = 333.2 kj/kg and δhv = 2257 kj/kg, respectively. use these values to estimate the total energy required to melt 100 kg of ice at 0 °c and boil off 40 kg of water at 100 °c. a) 239,028 kj b) 95,250 kj c) 185,500 kj d) 362,628 kj e) 123,600 kj

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of  vaporisation .

Sensible heat for water Q

For water

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

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