**Example #1:**determine the last temperature when 32.2 g that water in ~ 14.9 °C mixes with 32.2 grams that water at 46.8 °C.This is trouble 8a native Worksheet #2.First some discussion, climate the solution. Forgive me if the points seem obvious:1) The colder water will warmth up (heat power "flows" into it). The warmer water will certainly cool under (heat energy "flows" out of it).

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**2) The whole mixture will certainly wind up at the SAME**temperature. This is very, really important.

**3) The power which "flowed" the end (of the warmer water) equals the power which "flowed" in (to the colder water)This problem type becomes slightly harder if a phase adjust is involved. For this example, no step change. What that means is that only the details heat equation will certainly be involvedSolution an essential Number One:**We begin by calling the final, finishing temperature "x." keep in mind that BOTH water samples will certainly wind up in ~ the temperature we room calling "x." Also, make sure you understand that the "x" we are using IS not the Δt, yet the

**FINAL**temperature. This is what us are fixing for.The warmer water goes under from to 46.8 come x, so this way its Δt amounts to 46.8 − x. The colder water goes up in temperature, so its Δt amounts to x − 14.9.That last paragraph may be a little bit confusing, therefore let"s compare it come a number line:

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

**qlost = qgainHowever:q = (mass) (Δt) (Cp)So:(mass) (Δt) (Cp) = (mass) (Δt) (Cp)With qlost ~ above the left side and also qgain top top the ideal 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:**identify the final temperature once 45.0 g the water in ~ 20.0 °C mixes with 22.3 grams that water at 85.0 °C.

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**Solution:**We start by calling the final, ending temperature "x." store in mind that BOTH water samples will wind up in ~ the temperature we are calling "x." Also, make sure you know that the "x" we room using IS no the Δt, but the

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

**To compute the absolute distance, it"s the larger value minus the smaller sized value, so 85.0 come x is 85.0 − x and the street from x (the larger value) come 20.0 (the smaller sized value) is x − 20.0.The power amount going the end of the warm water is same to the energy amount going right into the cool water. This means:qlost = qgainSo, through substitution, we then have:(22.3) (85.0 − x) (4.184) = (45.0) (x − 20.0) (4.184)Solve for xExample #3:**recognize the last temperature once 30.0 g of water at 8.00 °C mixes with 60.0 grams that water in ~ 28.2 °C.

**Solution:**