The Attempt at a Solution for Jar of Tea Please Help Ice and a Jar of Tea

  • Thread starter Aikenfan
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In summary, the question asks for the mass of remaining ice in a jar of tea after adding 97.8g of ice at 0 deg Celsius to 187g of tea at an equilibrium temperature of 30.1 deg Celsius. Using the equation Q = mc (change in T), the attempt at a solution calculated 17.4g of remaining ice, but this is incorrect due to not accounting for the phase change of melting ice. A more accurate approach would be to use variables and assume that the remaining ice is still at 0 degrees. This would result in an equation of (97.8-x)(deltaT+mystery factor) = deltaT times mass of tea, where x represents the fraction of ice melted and
  • #1
Aikenfan
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Please Help Ice and a Jar of Tea

Homework Statement


Given; The specific heat of water is 4186 J/kg deg Celsius.
A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 30.1 deg Celsius. In an attempt to cool the liquid, which has a mass of 187g, 97.8g of ice at 0 deg Celsius is added. Assume the specific heat capacity of the tea to be that of pure liquid water. At the time at which the temperature of the tea is 27.3 deg Celsius, find the mass of the remaining ice in the jar. Answer in units of g


Homework Equations



Q = mc (change in T)

The Attempt at a Solution


Q = mc (change in T) = -mc (change in T)
(.187)(4186)(30.1-27.3) = -m (4186)(30.1)
= 17.4g

but i get the answer wrong when i put it into the computer...if anyone can help me with what i am doing wrong, it would be greatly appreciated!
 
Last edited:
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  • #2
There is a phase change which you have not accounted for. Even tho there is no change in temperature, "melting ice" or boiling water takes additional energy. the algebra is a little tricky therefore, but doable with careful use of variables, and the assumption I believe you need to make that the remaining ice is still at zero degrees. So one end of the eqn looks somthing like, assuming x to be the fraction melted and warmed:

(97.8-x)(deltaT {diff between 0 degrees and final}+mystery factor alluded to above)= deltaT (which is the diff beteen its initial temp and final temp) times mass of tea.

see if that helps.
 
Last edited:
  • #3


It seems like you are using the correct equation, Q=mc(change in T), to solve for the mass of remaining ice. However, it looks like you may have made a calculation error. When plugging in the values, be sure to use the correct signs for the temperatures (30.1 should be positive and 27.3 should be negative). Also, make sure to use the correct units for mass (kg) and specific heat (J/kg deg Celsius). Double check your calculations and make sure to include all units in your final answer. If you are still having trouble, try breaking down the problem into smaller steps and checking each step for accuracy.
 

1. What is a "Jar of Tea"?

A "Jar of Tea" is a beverage that is made by steeping tea leaves in hot water and then storing the resulting liquid in a jar.

2. How is a "Jar of Tea" different from regular tea?

A "Jar of Tea" is typically made in larger quantities and stored in a jar for later consumption. It can also be customized with different types of tea and additional ingredients for added flavor.

3. How long can a "Jar of Tea" be stored?

A "Jar of Tea" can be stored for up to 3 days in the refrigerator. It is recommended to consume it within this time frame for optimal freshness and taste.

4. Is a "Jar of Tea" healthy?

Yes, a "Jar of Tea" can be a healthy beverage depending on the type of tea and ingredients used. Green tea, for example, is known for its numerous health benefits.

5. Can you reuse the tea leaves from a "Jar of Tea"?

Yes, you can reuse the tea leaves from a "Jar of Tea" to make another batch. However, the resulting tea may be weaker in flavor compared to the first batch.

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