Calorimetry problem involving tea and ice

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Homework Help Overview

The problem involves a calorimetry scenario where hot tea is mixed with ice, and the goal is to determine the final equilibrium temperature of the mixture. The context includes specific heat capacities and heat of fusion relevant to the thermal interactions between the substances involved.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the expression for final temperature and question the specific heat value used for ice. There is an exploration of the relevance of specific heat in the context of melting ice versus changing its temperature.

Discussion Status

Some participants have provided guidance on the use of specific heat values, suggesting that the specific heat of water should be used instead of that of ice, given that the ice melts in the process. There is an ongoing exploration of the calculations and values being applied.

Contextual Notes

Participants note discrepancies in specific heat values for ice from various sources, indicating a potential source of confusion in the calculations. The problem assumes that all ice melts, which affects the relevance of specific heat in the calculations.

mcnivvitz
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The problem:

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 T_tea = 75.0 C. You then add 0.975 kg of ice taken from the freezer at a temperature of T_ice = 0 C. By the time the mix reaches equilibrium, all of the ice has melted. What is the final temperature T_f 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 is = 4190.
The heat of fusion of ice is = 3.33×105 .
The density of the tea is = 1.00 .

My attempt:

I wrote the following expression to find T_f:
T_f = (m_tea*c*T_tea + m_ice*c*T_ice - L_f*m_ice) / (cm_tea + cm_ice)

I'm pretty sure this is right, but when I plug in the variables, I get:

(1.5*4190*75 - .975*3.33e5) / (4190*1.5 + .975*2050) = 17.71 C

17 degrees seems like a reasonable answer but it's not right. What could I be doing wrong?
Thanks.
 
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Hello mcnivvitz,

Welcome to Physics Forums!
mcnivvitz said:
I wrote the following expression to find T_f:
T_f = (m_tea*c*T_tea + m_ice*c*T_ice - L_f*m_ice) / (cm_tea + cm_ice)

I'm pretty sure this is right, but when I plug in the variables, I get:

(1.5*4190*75 - .975*3.33e5) / (4190*1.5 + .975*2050) = 17.71 C
Where does the 2050 come from?
 
collinsmark said:
Hello mcnivvitz,

Welcome to Physics Forums!

Where does the 2050 come from?

2050 is the specific heat capacity of ice, although I'm not sure where I got that number... It seems every site I go to has a different value. Wikipedia is saying 2110 So I'll see if that makes a difference.

It gives me 17.6, which is still wrong. Any other suggestions?
 
mcnivvitz said:
2050 is the specific heat capacity of ice, although I'm not sure where I got that number... It seems every site I go to has a different value. Wikipedia is saying 2110 So I'll see if that makes a difference.

It gives me 17.6, which is still wrong. Any other suggestions?
Specific heat of ice. Hmmm. Now hold on a second.

The specific heat of ice might be useful if you are trying to find out how much energy it takes to change a block of ice by a given temperature. For example, if you wanted to change the temperature of a 1 kg block of ice from -20o C to -10o C, then the specific heat of ice would be useful. But the key point here is that it's only useful if the ice's temperature is changing and the block of ice remains ice.

Where is it in this problem where the temperature of the ice is changing (such that the ice remains in solid ice form, and is not a liquid or gas)? :wink:
 
I see. Use the specific heat of water instead of ice, since the ice melts. Thank you so much!
 

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