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Eugene S.
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Problem 104 Hard Difficulty

We mix 50.0 mL of ethanol (density = 0.789 g>mL) initially at 7.0 C with 50.0 mL of water (density = 1.0 g>mL) initially at 28.4 C in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture?

Answer

$$
21.6^{\circ} \mathrm{C}
$$

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Video Transcript

So in this problem, we are given that we have fifty milliliters of ethanol at seven point zero degrees Celsius and we have fifty milliliters of water at twenty eight point four degrees Celsius. And we mix these two together and we want to find the final temperature off the mixture. And so the way that we do that is the formula. Q. Gained equals Q lost. And so our cue gained corresponds to our ethanol and arc you lost corresponds to our water and so we can expand these out into ah, Mass and I'LL call it M E for massive ethanol times C e the specific heat of ethanol and then delta t e the change in temperature of the ethanol. And then we do the same for the water with our subscript w for water. And so now that we have that, we can plug in those values. And so for the mass of ethanol, we have fifty milliliters and we can multiply that by the density point seven eight nine grams per mil leader and that'LL give us are massing grams. We multiply by the specific heat two point four six Jules per gram degrees Celsius and then the change in temperature, which will be our final temperature, which is unknown minus or seven point zero degrees Celsius. And that is equal to the mass of water, which is fifty million leaders and also fifty grams. Because water's density is one Ah mo by the specific heat of water, which is four point one eight four Jules program degree Celsius and then now we take our temperature of our water and subtract our final temperature. And the reason the final temperature in different places here is because we want the each of these Delta teas to be positive so that our equation works. And since we know the final temperature will be somewhere in between these two by setting it up this way, we ensure that both of our delta teas are positive and so doing some math on this we end up with ninety seven point zero four seven temperature final minus six seventy nine point three to nine equals five nine for one point two eight minus two o nine point two temperature final. And so now we can combine like terms and we end up with three oh six point two four seventy F equals six six to zero point six o nine and that gives us that our final temperature is twenty one point six degrees Celsius and that is our final answer.

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