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• Calculate the heat energy required to convert completely 10 kg of water at 50°C into steam at 100°C, given that
the specific heat capacity of water is 4200 J/(kg°C) and the specific latent heat of vaporization of water
is 2260 kJ/kg.

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Starrysoul100

[tex]\bold{\huge{\green{\underline{ Solution}}}}[/tex]

Given :-

  • We have 10kg of water which is melting at the temperature of 50° C and 100° C
  • The specific heat capacity of water is 4200J/Kg
  • The specific latent heat of vaporisation of water is 2260 KJ/kg

Let's Begin :-

We have,

  • [tex]\sf{Mass = 10 kg }[/tex]
  • [tex]\sf{Temperature = 50° C and 100°C }[/tex]
  • [tex]\sf{ Heat \: Capacity = 4200J/kg}[/tex]
  • [tex]\sf{ L\: of \:vaporisation = 2260KJ/kg }[/tex]

We know that,

Heat required for temperature change in the same phase

[tex]\sf{\purple{Q = mcΔT }}[/tex]

Subsitute the required values,

[tex]\sf{Q = 10× 4200× ( 100 - 50) }[/tex]

[tex]\sf{Q = 10× 4200× 50 }[/tex]

[tex]\sf{Q = 4200 × 500 }[/tex]

[tex]\sf{Q = 2100000 J/Kg°C}[/tex]

[tex]\sf{\blue{Q = 2100 kJ/Kg°C}}[/tex]

Now,

Heat required for phase change at the same temperature

[tex]\sf{\orange{Q = mL }}[/tex]

  • L is the latent heat of vaporisation

Subsitute the required values,

[tex]\sf{Q = 10 × 2260 }[/tex]

[tex]\sf{\pink{Q = 22600 KJ/Kg}}[/tex]

Therefore,

Total heat energy required to convert 10kg of water

[tex]\sf{Q = mcΔT + mL}[/tex]

[tex]\sf{Q = 2100 + 22600}[/tex]

[tex]\sf{Q = 24700 KJ/Kg°C}[/tex]

[tex]\sf{Q = 24700000 J/Kg°C}[/tex]

[tex]\sf{\red{Q = 2.47 × 10^7 J/Kg°C}}[/tex]

Hence, The heat required to convert completely 10kg of water at 50° C into steam 100° C is 2.47 × 10^7 J/kg°C or 2.47 × 10^4 kJ/kg°C.