31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams of water at 24.5 degrees Celsius. If the final temperature reached in the calorimeter is 32.5 degrees Celsius, what is the specific heat of the unknown substance?

Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g).
@JustSaiyan

Question

31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams of water at 24.5 degrees Celsius. If the final temperature reached in the calorimeter is 32.5 degrees Celsius, what is the specific heat of the unknown substance?

Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g).
@JustSaiyan

sox2 · Answer

not looking for help, just direct answers

sox2 · Answer

WELL NEVER FCKING MIND BECAUSE MY EXAM JUST CLOSED OUT AND EXPIRED

sox2 · Answer

GOD ***ING DAMNIT

JustSaiyan · Answer

msDT = msDT

m = mass 
s = specific heat 
DT = change in temperature

(31.5)(x)(102.4-32.5) = (103.5)(4.18)(32.5-24.5) 
2201.85x = 3464.352 
x = 1.57 J/(° C × g).

JustSaiyan · Answer

Sorry bud. Took me too long. Just take this medal.

sox2 · Answer

i got the exam unlocked again thank god