# 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.
The text contains a physics problem related to heat transfer and phase change, specifically about mixing hot tea with ice. The problem is broken down into two parts, each with a different initial temperature for the tea, and the solutions for both parts are provided.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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.
user
The text contains a physics problem related to heat transfer and phase change, specifically about mixing hot tea with ice. The problem is broken down into two parts, each with a different initial temperature for the tea, and the solutions for both parts are provided.

TYPE: SINGLE_QUESTION_WITH_ANSWER