text: # Work done in converting 1g of ice at -10°C into steam at 100°C is 3.04 kJ 6.05 kJ 0.721 kJ 0.616 kJ
  
  Arun
  25750 Points
  4 years ago
  Heat Energy Required to convert 1 g ice into 1 g water = mL
  [L is the Latent heat of melting of Ice.]
  = 1g × 336 J/g.
  = 336 J.
  Heat Energy Required to raise the temperature of the Water from 0°  C to 100°C = mcΔt
  [c is the specific heat capacity of water = 4200 J/kg°C.]
  = 1/1000 × 4200 × 100
  = 420 J.
  Heat Energy Required to convert water at 100° C into steam = mL
  [L is the latent heat of Vaporization of Water = 2280 J/g.]
  = 1 × 2260 J/g.
  = 2260 J
  Total Energy Required in this Thermodynamic Process = 336 + 420 + 2260
  = 3016 J.
  Hope it helps.
explanation: The document contains a physics problem related to calculating the heat energy required to convert 1 gram of ice at -10°C into steam at 100°C. The problem is followed by a detailed step-by-step solution provided by Arun, which includes the calculation of the energy required for each phase change and temperature increase. The final answer is given as 3016 J.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Work done in converting 1g of ice at -10°C into steam at 100°C is 3.04 kJ 6.05 kJ 0.721 kJ 0.616 kJ
  
  Arun
  25750 Points
  4 years ago
  Heat Energy Required to convert 1 g ice into 1 g water = mL
  [L is the Latent heat of melting of Ice.]
  = 1g × 336 J/g.
  = 336 J.
  Heat Energy Required to raise the temperature of the Water from 0°  C to 100°C = mcΔt
  [c is the specific heat capacity of water = 4200 J/kg°C.]
  = 1/1000 × 4200 × 100
  = 420 J.
  Heat Energy Required to convert water at 100° C into steam = mL
  [L is the latent heat of Vaporization of Water = 2280 J/g.]
  = 1 × 2260 J/g.
  = 2260 J
  Total Energy Required in this Thermodynamic Process = 336 + 420 + 2260
  = 3016 J.
  Hope it helps.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a physics problem related to calculating the heat energy required to convert 1 gram of ice at -10°C into steam at 100°C. The problem is followed by a detailed step-by-step solution provided by Arun, which includes the calculation of the energy required for each phase change and temperature increase. The final answer is given as 3016 J.
  
  Label: question-and-answer