If 50.0 mL of ethanol (density = 0.789 g>mL) initially at 7.0 °C is mixed with 50.0 mL of water (density = 1.0 g/mL) initially at 28.4 °C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?

Video transcript

All right. Hi, everyone. So this question says that if 50.0 mL of ethanol with the density of 0.789 g per milliliter initially at 7.0 °C is mixed with 50.0 mL of water with the density equal to 1.0 g per milliliter initially at 28.4 °C in an insulated beaker and assuming that no heat is lost. What is the final temperature of the mixture? Option A says 28.2 °C. Option B says 46.1 °C. Option C says 35.4 °C and option D says 21.7 °C. So recall that this question in particular is discussing thermal equilibrium between the two substances that are being mixed together. In this case, thermal equilibrium will be reached between ethanol and water in the solution. Now, because no heat is lost during this process, right, the heat is going to be transferred between each substance. So as heat get lost, buy water, it is going to be absorbed, buy ethanol. This ensures that no heat escapes the system. Now, recall also that when heat is absorbed, the value of Q is positive. Whereas if it loses heat, the value of Q is negative. So this means that negative Q for water is equal to positive Q or ethanol. And so we're going to use this relationship to find the final temperature of the system because the final temperature of ethanol at thermal equilibrium is equal to the final temperature of water. And the final temperature of the system overall recall that Q is equal to MC delta T where M is the mass C is the specific heat of the substance. And delta T refers to the change in temperature. Now to be a bit more specific Q can also be written as MC multiplied by TF subtracted by T I where TF is the final temperature and T I is the initial temperature. So in this case, right, we can find the value for the Q of water or in other words, negative MC delta T or water and equal this to the MC delta T of ethanol. So let's go ahead and substitute our values for each one starting off with water. The mass of water in the sample is equal to the volume of water multiplied by its density. So if there are 50 mL of water and the density is 1.0 g per milliliter, the mass of water is going to be 50.0 g. So that's negative 50.0 g multiplied by the specific heat of water, which if you recall is 4.18 joules per gram degree Celsius. And this is multiplied by the difference between the final temperature, which is what we're solving for, subtracted by the initial temperature of water, which was 28.4 °C. So this is going to be equal to the value of Q four ethanol. But for ethanol, we, we are going to need to calculate its mass. So as discussed previously, we're going to take the volume of ethanol and multiply that by its density. So the mass of ethanol is equal to 50.0 mL multiplied by 0.789 g per milliliter. And this yields a mass of 39.45 g or ethanol. So that is what I'm going to plug in here or the mass of ethanol on the right side of this expression. So thats 39.45 g multiplied by the specific heat of ethanol, which is 2.42 joules per gram degree Celsius multiplied by TF subtracted by 7.0 °C, which was the initial temperature of ethanol. So my objective here, at least at this point is to simplify this expression as best as I can, right. So I'm going to start by multiplying the mass and the specific heat on both sides of the equation. So in the case of water, that would be negative 50.0 g including the negative sign here, multiplied by 4.18 joules per gram degree Celsius, combining these together, my product is negative 209 jules per degree Celsius. And so this is multiplied by the final temperature subtracted by 28.4 °C. So now let's go ahead and do the same thing or ethanol. I'm going to take the massive ethanol that's 39.45 g and multiply it by the specific heat which is 2.42 joules per gram degree Celsius. This gives me 95.469 joules per degree Celsius. And this is multiplied by TF subtracted by 7.0 °C. So at this point, right, we can use the distributive property on both sides of the equation. The products that we previously received can be distributed into the expression or delta T. In other words, TF subtracted by T initial. So on the left side here that becomes negative 209 joules per degree Celsius multiplied by TF subtracted by negative 209 joules per degree Celsius multiplied by negative 28.4 °C. This actually yields a positive 5935.6 jules. And so this is equal to on the right side, 95.469 joules per degree Celsius multiplied by TF subtracted by 95.469 joules per degree Celsius multiplied by negative 7.0 °C. And so this yields 668 0.283 jules or negative 668.283 jules. All right. So at this point, our next step is to combine like terms. So what I'm going to do here is add both sides of the equation by 209 joules per degree Celsius multiplied by TF And at the same time, I'm going to add both sides by 668.283 jules. This ensures that like terms have moved to one side of the equation, right? So 95 0.469 joules per degree Celsius multiplied by TF added to 209 joules per degree Celsius multiplied by TF yields 304 point 469 Jules per degree Celsius multiplied by TF. And this equals the sum of both values expressed in jewels. So that's 5935.6 added to 668.283. So that equals 6603 0.883 jules. And so the final temperature is equal to 6603 0.883 Jules divided by 304.469 joules per degree Celsius. And so after rounding to three significant figures, our answer becomes 21.7 °C. And if I scroll up here, this corresponds with option D in the multiple choice and there you have it. So if you watch this video all the way to the end. Thank you so very much for watching. I appreciate that. And I hope you found this helpful.