On a hot summer day, you decide to make some iced tea. First, you brew 1.50 L of
ID: 1350156 • Letter: O
Question
On a hot summer day, you decide to make some iced tea. First, you brew 1.50 L of hot tea and leave it to steep until it has reached a temperature of Ttea = 75.0 C. You then add 0.975 kg of ice taken from the freezer at a temperature of Tice = 0C. By the time the mix reaches equilibrium, all of the ice has melted. What is the final temperature Tf of the mixture?
For the purposes of this problem, assume that the tea has the same thermodynamic properties as plain water.
The specific heat of water isc = 4190 J/kgC .
The heat of fusion of ice isLf = 3.33×105 J/kg .
The density of the tea istea = 1.00 kg/L .
Question:
You found that the final temperature of the iced tea is between 0 C and 75.0 C, as the strategy indicated it should be. Evaluate the work you did to find the final temperature by considering how it would change if the initial conditions were varied.
You decide to make several more batches of iced tea. The conditions under which you make these batches are identical except that for each batch one initial condition is varied. Determine the impact that each of these changes will have on the final temperature of the iced tea.
Decrease the volume of the tea.
Decrease the mass of ice.
Increase the initial temperature of the tea.
Decrease the initial temperature of the ice.
Sort these changes to the initial conditions of the system based on whether they would cause the final temperature to be smaller than or greater than the temperature of the system found in Part C.
Explanation / Answer
Given
Volume of tea V Tea = 1.50 L =
Temperature of Tea Ttea = 75o C
Temperature of ice Tice = 0o C
Mass of the ice mice =0.975 Kg
The specific heat of water is c = 4190 J/kgC
The heat of fusion of ice is Lf = 3.33×105 J/kg
The density of the tea is tea = 1.00 kg/L
Solution
Volume of the tea = 1.5 L
Mass of the tea = Vtea x tea
Mtea = 1.50 x 1.00
Mtea = 1.50 Kg
Heat lost by tea to reach temperature Tf = Heat absorbed by ice ( at 0oC) to melt into water (0oC) + Heat absorbed by water (at 0oC) to rach the final temperature Tf
MteaC(Ttea – Tf) = MiceLf + MiceC(Tf –Tice)
1.5 x 4190 x (75-Tf) = 0.975 x 3.33 x 105 + 0.975 x 4190 ( Tf)
471375 – 6285Tf = 324675 + 4085.25 Tf
Tf =146700 / 10370.25
Tf = 14.15 OC
MteaC(Ttea – Tf) = MiceL + MiceC(Tf –Tice)
MteaCTtea - MteaCTf = MiceL + MiceCTf - MiceCTice
MteaCTtea - MiceL + MiceCTice = MteaCTf + MiceCTf
MteaCTtea - Mice(L – CTice) = (MteaC+ MiceC)Tf
Tf = {MteaCTtea - Mice(L – CTice)} /(MteaC+ MiceC)
Decrease the volume of the tea.
The final temperature decreases since heat needed to be lost increases
Verification
Let us decrease the volume of tea to 1L
MteaC(Ttea – Tf) = MiceLf + MiceC(Tf –Tice)
1 x 4190 x (75-Tf) = 0.975 x 3.33 x 105 + 0.975 x 4190 ( Tf)
314250 – 4190Tf = 324675 + 4085.25 Tf
-10425 = 8272.25Tf
Tf = -1.26 oC
So the final temperature decreases
Decrease the mass of ice.
The final temperature increases since the heat needed to be lost decreases
Verification
Let us decrease the mass of the ice to 0.875 Kg
MteaC(Ttea – Tf) = MiceLf + MiceC(Tf –Tice)
471375 – 6285Tf = 291375 + 3666.25Tf
Tf = 180000/9951.25 = 18.09oC
Increase the initial temperature of the tea.
The final temperature increases since the Tf is directly proportional to Ttea
Tf = {MteaCTtea - Mice(L – CTice)} /(MteaC+ MiceC)
Tf Ttea
Verification
increasing Ttea to 85oC
MteaC(Ttea – Tf) = MiceL + MiceC(Tf –Tice)
1.5 x 4190 x (85-Tf) = 324675 + 4085.25 Tf
502800 – 6285Tf = 324675 + 4085.25 Tf
Tf = 17.1oC
Decrease the initial temperature of the ice
The final temperature decreases since the Tf is directly proportional to Tice
Tf = {MteaCTtea - Mice(L – CTice)} /(MteaC+ MiceC)
Tf Tice
Verification
decreasing Tice to -10oC3
MteaC(Ttea – Tf) = MiceL + MiceC(Tf –Tice)
471375 – 6285Tf = 324675 + 4085.25 Tf + 4075.25
150775.25 = 10370.25 Tf
Tf = 13.75oC
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.