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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.

1 Answer

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\bold{\huge{\green{\underline{ Solution}}}}

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,


  • \sf{Mass = 10 kg }

  • \sf{Temperature = 50° C and 100°C }

  • \sf{ Heat \: Capacity = 4200J/kg}

  • \sf{ L\: of \:vaporisation = 2260KJ/kg }

We know that,

Heat required for temperature change in the same phase


\sf{\purple{Q = mcΔT }}

Subsitute the required values,


\sf{Q = 10× 4200× ( 100 - 50) }


\sf{Q = 10× 4200× 50 }


\sf{Q = 4200 × 500 }


\sf{Q = 2100000 J/Kg°C}


\sf{\blue{Q = 2100 kJ/Kg°C}}

Now,

Heat required for phase change at the same temperature


\sf{\orange{Q = mL }}

  • L is the latent heat of vaporisation

Subsitute the required values,


\sf{Q = 10 × 2260 }


\sf{\pink{Q = 22600 KJ/Kg}}

Therefore,

Total heat energy required to convert 10kg of water


\sf{Q = mcΔT + mL}


\sf{Q = 2100 + 22600}


\sf{Q = 24700 KJ/Kg°C}


\sf{Q = 24700000 J/Kg°C}


\sf{\red{Q = 2.47 × 10^7 J/Kg°C}}

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.

User Kristoffer
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