heat lost by hot metal + heat gained by cooler water = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial)] = 0
Solve for specific heat metal. You have all of the other items.
An insulated beaker contains 250.0 grams of water at 25C. Exactly 41.6 g of a metal at 100.0C was dropped in the beaker. The final temperature of the water was 26.4C. Assuming that no heat is lost in any other way, calculate the specific heat of the metal.
4 answers
did not help
answer choices:
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
answer choices:
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
Read my response at your later post. You missed something.
[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?