**Example #1:**recognize the final temperature once 32.2 g of water at 14.9 °C mixes v 32.2 grams that water in ~ 46.8 °C.This is difficulty 8a from Worksheet #2.First some discussion, then the solution. Forgive me if the points it seems ~ obvious:1) The colder water will heat up (heat power "flows" right into it). The warmer water will certainly cool down (heat energy "flows" the end of it).

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**2) The whole mixture will wind up in ~ the SAME**temperature. This is very, very important.

**3) The power which "flowed" out (of the warmer water) equals the power which "flowed" in (to the colder water)This problem kind becomes slightly harder if a phase change is involved. For this example, no phase change. What that method is that only the details heat equation will certainly be involvedSolution vital Number One:**We begin by call the final, ending temperature "x." save in mind that BOTH water samples will wind up at the temperature we are calling "x." Also, make certain you know that the "x" we room using IS no the Δt, yet the

**FINAL**temperature. This is what us are fixing for.The warmer water goes down from to 46.8 come x, so this way its Δt equals 46.8 − x. The chillier water goes increase in temperature, therefore its Δt equals x − 14.9.That last paragraph might be a bit confusing, for this reason let"s compare it come a number line:

**To compute the absolute distance, it"s the bigger value minus the smaller sized value, for this reason 46.8 to x is 46.8 − x and also the distance from x to 14.9 is x − 14.9.These two distances on the number line stand for our two Δt values:a) the Δt of the warmer water is 46.8 minus xb) the Δt the the cooler water is x minus 14.9Solution crucial Number Two:**the power amount going out of the heat water is equal to the power amount going right into the cool water. This means:

**qlost = qgainHowever:q = (mass) (Δt) (Cp)So:(mass) (Δt) (Cp) = (mass) (Δt) (Cp)With qlost top top the left side and also qgain on the best side.Substituting values into the above, us then have:(32.2) (46.8 − x)(4.184) = (32.2) (x − 14.9) (4.184)Solve because that xExample #2:**determine the final temperature as soon as 45.0 g the water in ~ 20.0 °C mixes through 22.3 grams that water in ~ 85.0 °C.

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**Solution:**We begin by calling the final, finishing temperature "x." save in mind the BOTH water samples will certainly wind up in ~ the temperature we are calling "x." Also, make sure you recognize that the "x" we room using IS no the Δt, however the

**FINAL**temperature. This is what us are fixing for.The warmer water goes down from to 85.0 come x, therefore this method its Δt amounts to 85.0 minus x. The colder water goes increase in temperature (from 20.0 come the finishing temperature), therefore its Δt equates to x minus 14.9.That last paragraph may be a little bit confusing, so let"s to compare it come a number line:

**To compute the absolute distance, it"s the bigger value minus the smaller value, therefore 85.0 come x is 85.0 − x and also the distance from x (the bigger value) come 20.0 (the smaller sized value) is x − 20.0.The energy amount going out of the heat water is same to the power amount going into the cool water. This means:qlost = qgainSo, by substitution, we then have:(22.3) (85.0 − x) (4.184) = (45.0) (x − 20.0) (4.184)Solve because that xExample #3:**identify the last temperature once 30.0 g the water at 8.00 °C mixes v 60.0 grams of water in ~ 28.2 °C.

**Solution:**