In a physics lab experiment, a student immersed 205 one-cent coins (each having a mass of 3.00 g *.003kg) in boiling water. After they reached thermal equilibrium, she quickly fished them out and dropped them into 0.244 kg of water at 20.0 C in an insulated container of negligible mass.
What was the final temperature of the coins? [One-cent coins are made of a metal alloy - mostly zinc - with a specific heat capacity of 390 J/(kg*K).]
I know that:
Mass,coins=.615 kg
Mass,water=.244 kg
c,coin=390 J/(kg*K)
c,water=418.6 J/(kg*K)
Ti, coins= 100C, 373.15K
Ti, water= 20C, 293.15K
The problem asks for the Tf of the coins, but I'm still missing the Tf of the water...I'm thinking it has something to do with the equilibrium point, but I dont' know that that is!
Heat lost by the coins + heat gained by the water = 0
mass x specific heat x (Tf - Ti) + mass x specific heat x (Tf - Ti) = 0
You posted this yesterday and I gave you the formula, which you have copied.
I think you have missed the point. For the first
mass x specific heat x (Tf-Ti) , you have mass coins, spcific heat of coins, and Ti of coins.
For the second mass x specific heat x (Tf-Ti) you have
mass water, specific heat water, and Ti of water. That leaves only one unknown, namely, Tfinal which is what you are looking for. Solve the above for Tf. BIG Hint: Won't the Tf of the water be the same as the Tf of the coins. You're putting hot coins in cool water and both will come to some final T. The coins will lose heat, the water will gain heat, and the process of losing and gaining will stop WHEN BOTH ARE THE SAME. That will be Tfinal. Solve for Tfinal.
Post your work if you get stuck.
ok, i thought i had an answer, but it wasn't correct...here was my work...
(.615)(390)(T-373.15)=(.244)(418.6)(T-293.15)=0
T=349.2571663 -> C=76.1071663
When I used that T in the equation it worked... =(
It's set up right. All you need to do is to solve for T. You can make the numbers a little smaller if you use C instead of K. It's the difference you want and the difference is the same in K or C.
To solve this problem, you can create an equation based on the principle of energy conservation. The heat lost by the coins should be equal to the heat gained by the water.
The equation can be written as:
(mass of coins * specific heat capacity of coins * (Tf of coins - Ti of coins)) + (mass of water * specific heat capacity of water * (Tf of water - Ti of water)) = 0
Let's substitute the known values into the equation:
(0.615 kg * 390 J/(kg*K) * (Tf - 373.15 K)) + (0.244 kg * 418.6 J/(kg*K) * (Tf - 293.15 K)) = 0
Simplifying the equation gives:
(240.435 Tf - 222365.81775) + (102.5316 Tf - 71547.3266) = 0
Combine like terms:
343.9666 Tf - 293913.14435 = 0
Now, solve for Tf by isolating it:
343.9666 Tf = 293913.14435
Tf = 293913.14435 / 343.9666
Tf ≈ 854.0 K
So, the final temperature of the coins is approximately 854 Kelvin.