Question

0.500 kg of ethyl alcohol at 33.0°C

are added to 0.500 kg of water at
22.0°C. What is their equilibrium
temperature?​

Answer 1

26.06

Step-by-step explanation:

One thing to note first is that, after the ethyl alcohol and water is mixed, the heat energy (Q) is equal for both of them. So:

m c ΔT (water) = m c ΔT (ethyl)

Second thing to note is that ΔT is just the absolute value of the final temperature of the solution minus the initial temperature of the ethyl, or water. So the previous equation becomes:

m c (Tfin - Tinit) (water) = m c (Tfin - Tinit) (ethyl)

And since we know the mass and specific heat of the water and ethyl alcohol, we can substitute those in:

2093(Tfin - 22) = 1225(33 - Tfin)

(I wrote 33 - Tfin for ethyl because ΔT should be a positive number in this case)

Now finally we can expand out the parenthesis and use a little algebra to solve:

2093Tfin - 46046 = 40425 - 1225Tfin

Now use algebra to separate the Tfin coefficients onto one side and everything else on the other, and we get:

3318Tfin = 86471

And when you divide both sides be 3318, you get your answer of:

Tfin = 26.06118 or 26.1 for significant figures

Hope that helps!