Question

• 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.

Answer 1

`\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.