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.
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
2 answers
[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?
6.24m+113.022=0
62.4m=-113.022
m=113.022/62.4
m=1.81125
not an answer choice?