Achieve Hw (Week 3 and 4)#9
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 35
- Joined: Mon Jan 09, 2023 9:08 am
Achieve Hw (Week 3 and 4)#9
How would we know which equation to use to find the final temperature for this question?
-
- Posts: 36
- Joined: Mon Jan 09, 2023 9:27 am
Re: Achieve Hw (Week 3 and 4)#9
For this question we assume no heat is lost to the surroundings, so the q gained for the colder water is equal to the q lost for the warmer water (q=-q). We know that our equation for q is q=mCt (with m being grams, C being specific heat, and t being the change in temperature). We will be solving for the delta t value, which tells us the final temp.
Re: Achieve Hw (Week 3 and 4)#9
For a visual, it would be m*C*(Tf - Ti) = -[m*C*(Tf-Ti)]. As mentioned, solving for Tf and with C equal to 4.18.
-
- Posts: 41
- Joined: Mon Jan 09, 2023 2:29 am
Re: Achieve Hw (Week 3 and 4)#9
yesmal1J wrote:For a visual, it would be m*C*(Tf - Ti) = -[m*C*(Tf-Ti)]. As mentioned, solving for Tf and with C equal to 4.18.
I've been trying this one for a while, but haven't gotten the right answer. What steps did you take to get Tf? I tried solving for it and kept getting numbers that were way off.
Re: Achieve Hw (Week 3 and 4)#9
So the formula, m*C*(Tf - Ti) = -(m*C*(Tf-Ti)), is basically cold water=-hot water. My question was:
If you combine 290.0 mL of water at 25.00 ∘C and 130.0 mL of water at 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
So I did (290.0g)(4.184 J/g*∘C)(Tfinal - 25.00 ∘C) = (-130.0g)(4.184J/g*∘C)(Tfinal-95.00C) which gets you a final answer of 46.67 ∘C. Hope that helps!
If you combine 290.0 mL of water at 25.00 ∘C and 130.0 mL of water at 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
So I did (290.0g)(4.184 J/g*∘C)(Tfinal - 25.00 ∘C) = (-130.0g)(4.184J/g*∘C)(Tfinal-95.00C) which gets you a final answer of 46.67 ∘C. Hope that helps!
-
- Posts: 49
- Joined: Mon Jan 09, 2023 8:27 am
Re: Achieve Hw (Week 3 and 4)#9
I struggled with this too! I think the main thing is to set the two m(c)(deltaT) equations equal to each other and make sure to make one side of the equation negative. Also when solving, try to remember to follow order of operations and distribute the the values inside the parentheses. Sometimes there's just small algebraic errors that get you to the wrong answers.
-
- Posts: 39
- Joined: Mon Jan 09, 2023 9:00 am
Re: Achieve Hw (Week 3 and 4)#9
Hello. Because we know q=mC(Tf-Ti) and that heat is neither entering or being removed from the system, we are able to set m_coldC(Tf-Ti) = -m_hotC(Tf-Ti). Since cold water is gaining heat, the equation for cold water is positive. With that, the equation for hot water is negative because it is losing heat. As you can see, heat is being transferred between water of different temperatures, thus no heat is added or removed. Hope this helps!
Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”
Who is online
Users browsing this forum: No registered users and 7 guests