text: # Homework Help: Making iced tea
  
  1. Dec 8, 2012
  
  ### phunphysics2
  
  a person makes a quantity of iced tea by mixing 500 g of hot tea (essentially water) with an equal mass of ice at its melting point. if the initial hot tea is at room temperature of (a) 90 degrees celsius and (b) 70 degrees celsius, what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.
  
  My attempt
  
  Latent heat of melting - 334 kJ/kg=L
  s=Specific heat water - 4.187 kJ/kgK
  from calorimetry principle,
  heat rejected by tea=heat used to melt ice
  a)Final temp=t degree
  all ice melts
  heat rejected=ms(90-t)=.5*4.187*(90-t)
  heat tken in ice=mL+mst=.5*334+.5*4.187*t
  so .5*4.187*(90-t)=.5*334+.5*4.187*t
  temp=5.11
  
  b)let m ice melts
  final temp=0
  heat rejected=ms(70-0)=.5*4.187*(70-0)=146.545
  heat taken in ice=mL=m*334
  so 146.545=m*334
  so m=.439 g
  so ice remains=.5-.439=.061
  
  ans :
  a)final temp=5.11 degree c ; no ice remains
  b)Final temp=0, .061 kg or 61.2 gram ice remain
  
  Can someone please tell me why in Part A 90 is being used as the final temperature and t is used as the initial temperature? Thank you! I am really confused about that part
  
  2. Dec 8, 2012
  
  ### TSny
  
  Heat rejected is the negative of heat gained. To take care of the negative, the initial and final temperatures are switched in calculating the change in temperature for the heat rejected. That way, you get a positive number for the heat rejected.
explanation: The text extract is a homework help request and response regarding a physics problem about making iced tea. The user, phunphysics2, presents a problem involving the mixing of hot tea and ice, and provides their attempt at solving it. They then ask for clarification on a specific part of their solution. The response from TSny explains the reasoning behind the temperature terms in the equation.
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distilabel_metadata.content: # Homework Help: Making iced tea
  
  1. Dec 8, 2012
  
  ### phunphysics2
  
  a person makes a quantity of iced tea by mixing 500 g of hot tea (essentially water) with an equal mass of ice at its melting point. if the initial hot tea is at room temperature of (a) 90 degrees celsius and (b) 70 degrees celsius, what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.
  
  My attempt
  
  Latent heat of melting - 334 kJ/kg=L
  s=Specific heat water - 4.187 kJ/kgK
  from calorimetry principle,
  heat rejected by tea=heat used to melt ice
  a)Final temp=t degree
  all ice melts
  heat rejected=ms(90-t)=.5*4.187*(90-t)
  heat tken in ice=mL+mst=.5*334+.5*4.187*t
  so .5*4.187*(90-t)=.5*334+.5*4.187*t
  temp=5.11
  
  b)let m ice melts
  final temp=0
  heat rejected=ms(70-0)=.5*4.187*(70-0)=146.545
  heat taken in ice=mL=m*334
  so 146.545=m*334
  so m=.439 g
  so ice remains=.5-.439=.061
  
  ans :
  a)final temp=5.11 degree c ; no ice remains
  b)Final temp=0, .061 kg or 61.2 gram ice remain
  
  Can someone please tell me why in Part A 90 is being used as the final temperature and t is used as the initial temperature? Thank you! I am really confused about that part
  
  2. Dec 8, 2012
  
  ### TSny
  
  Heat rejected is the negative of heat gained. To take care of the negative, the initial and final temperatures are switched in calculating the change in temperature for the heat rejected. That way, you get a positive number for the heat rejected.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a homework help request and response regarding a physics problem about making iced tea. The user, phunphysics2, presents a problem involving the mixing of hot tea and ice, and provides their attempt at solving it. They then ask for clarification on a specific part of their solution. The response from TSny explains the reasoning behind the temperature terms in the equation.
  
  Label: question-and-answer