An insulated beaker contains 250.0 grams of water at 25oC. Exactly 41.6 g of a
metal at 100.0oC was dropped in the beaker. The final temperature of the water
was 26.4oC. Assuming that no heat is lost in any other way, calculate the specific heat of the metal.
a. 0.159 J g-1 K-1
b. 0.478 J g-1 K-1
c. 0.0503 J g-1 K-1
d. 2.09 J g-1 K-1
i have no idea how to do this?
and yes i saw the other post but it did not help a bit
chemistry - DrBob222, Sunday, January 22, 2012 at 3:25pm
What do you mean it didn't help? Then you didn't read it (or try it). Specific heat of the metal is the ONLY unknown in that equation and if you solve for it you get one of the answers listed. I just solved it so I know that's true.
Show your work and I'll be able to tell you what you're missing.
chemistry - utna, Sunday, January 22, 2012 at 9:52pm
[41.6xmx1.5]+[18x4.186x1.5]=0
6.24m+113.022=0
62.4m=-113.022
m=113.022/62.4
m=1.81125
not an answer choice?
If we take the specific heat of water as 4.18 j/g/K
Do this by heat balance
Heat before
water 250x(25+273)x4.18
metal 41.6x(100+273)xk
after
water 250x(26.4+273)x4.18
metal 41.6x(26.4+273)xk
so
311410J + 15516.8k J = 312873J + 12455k J
solve for k
I got
0.4778 J/g/k
so B
Thank you soo much!
Here is your problem pasted from above. Actually, more than one problem.
41.6 is right.
x for specific heat right.
The 1.5 is supposed to be Tf-Ti which is 26.4-100 = -73.6.
Where you have 18 (where did you get 18?) is supposed to be 250.
4.186 is right
1.5 is supposed to be Tf-Ti which is 26.4-25 = 1.4. So if you solve all of that it looks like this.
[41.6*x*(-73.6)] + (250*4.18*1.4) = 0
-3061.76x + 1463 = 0
x = specific heat = 1463/3061.76 = 0.4778 which rounds to 0.478 J/g*C.
And that is one of the answer choices.
[41.6xmx1.5]+[18x4.186x1.5]=0
6.24m+113.022=0
62.4m=-113.022
m=113.022/62.4
m=1.81125
not an answer choice?
To solve this problem, we can use the principle of energy conservation. The heat lost by the metal when it cools down will be equal to the heat gained by the water when it warms up. The equation we can use to represent this is:
Qlost = Qgained
The heat lost by the metal can be calculated using the equation:
Qlost = m * c * ΔT
Where:
m = mass of the metal (41.6 g)
c = specific heat of the metal (unknown)
ΔT = change in temperature of the metal (100.0oC - final temperature)
Substituting the values given:
Qlost = 41.6 g * c * (100.0oC - 26.4oC)
The heat gained by the water can be calculated using the equation:
Qgained = m * c * ΔT
Where:
m = mass of the water (250.0 g)
c = specific heat of water (4.186 J g-1 K-1)
ΔT = change in temperature of the water (final temperature - initial temperature)
Substituting the values given:
Qgained = 250.0 g * 4.186 J g-1 K-1 * (26.4oC - 25oC)
Since Qlost = Qgained, we can set the two equations equal to each other and solve for c:
41.6 g * c * (100.0oC - 26.4oC) = 250.0 g * 4.186 J g-1 K-1 * (26.4oC - 25oC)
41.6 g * c * 73.6oC = 250.0 g * 4.186 J g-1 K-1 * 1.4oC
Simplifying the equation:
c = (250.0 g * 4.186 J g-1 K-1 * 1.4oC) / (41.6 g * 73.6oC)
c = 0.478 J g-1 K-1
Therefore, the specific heat of the metal is 0.478 J g-1 K-1, which corresponds to option (b).