A person makes a quantity of iced tea by mixing 730 g of hot tea (essentially wa

Question

A person makes a quantity of iced tea by mixing 730 g of hot tea (essentially water) with an equal mass of ice at its melting point. Assume the mixture has negligible energy exchanges with its environment. (a) If the tea's initial temperature is Ti = 94°C, when thermal equilibrium is reached what are the mixture's temperature Tf and (b) the remaining mass mf of ice? If Ti = 70°C, when thermal equilibrium is reached what are (c) Tf and (d) mf? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg.
Give the temperatures in Celsius and the masses in grams.

Explanation / Answer

Okay, first things first, you need to convert g into kg (Because specific heat and latent heat of fusion use kg in their units), making 730g into .73kg.

Then set up your equation. The main idea behind this kind of a problem is that the heat gained should always equal the heat lost from the system, and the equation is literally just that.

mLf (of the ice melting) + cmT (of the ice rising in temperature) = cmT (of the tea lowering in temperature)

Thus -- (.73kg)(3.33*10^5 J/kg) + (.73kg)(4186 J/kg.K)(Tf) = (.73kg)(4186)(94-Tf) -- and then solve for Tf, the final temperature at which the system will be at equalibrium, to get part A). [The T for the ice rising in temperature is only Tf because the ice is starting at 0 degrees, its melting point]

Check my work, but my answer was: A) 7.2 degrees Celsius.

For B) the answer is 0g, because all the ice will be melted when the ice and tea reach equilibrium.

The wording in your question for C and D confused me somewhat but what I think you are asking is to find the new Tf of equilibrium if the initial temperature of the tea is lowered. Then, D, if the initial temperature of the tea starts at 70 (instead of 94) will there be any ice left over and, if so, how much. If I've misinterpreted your question, please let me know and I'll be happy to edit my response to be more help. :)

If that's what you're looking for, however...

For C) you are, first, going to use the exact same equation that you just used a moment ago only, instead, inputting 70 degrees as your Ti as opposed to 94.

You should be setting it up as: (.73kg)(3.33*10^5) + (.73kg)(4186)(Tf) = (.73kg)(4186)(70-Tf)

Then...solve for Tf. Again.

What I got was a negative number, C) -4.75 degrees Celsius.

Thus giving us our first hint toward solving problem D. If your final temperature is below 0 degrees then, at some point, the ice will stop melting and, yes, there will be some left over. In that case, the next step is just another version of the same equation all over again.

This time, plug in your new Ti (70) and Tf(-4.75) and solve, instead, for the mass of ice. It should go as follows: (m)(3.33*10^5) + (m)(4186)(-4.75) = (.73kg)(4186)[70-(-4.75)] And then solve for m.

What I ended up getting was .729503 kg as my mass. But REMEMBER this equation only solves for how much of the ice has melted, not how much of the ice is left over.

To find D): .73-.729503 and you get .000497kg of ice left over.

Convert that back to grams from the beginning of your question and you get D) .497g of ice left over in the tea, which- is hardly any at all.

I've never seen a thermodynamics problem like this with such small values for answers, so it had me all paranoid. To the best of my knowledge, though, that should be it. :) Hope that helps.

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