35.5 grams of an unknown substance is heated to 103.0 degrees Celsius and then placed into a calorimeter containing 100.0 grams of water at 24.0 degrees Celsius. If the final temperature reached in the calorimeter is 29.5 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).

Question

35.5 grams of an unknown substance is heated to 103.0 degrees Celsius and then placed into a calorimeter containing 100.0 grams of water at 24.0 degrees Celsius. If the final temperature reached in the calorimeter is 29.5 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).

hayhayz · Answer

@sweetburger

aaronq · Answer

Use the calorimetry equation and the data regarding the calorimeter to find the heat absorbed by the water.

This same quantity of heat was lost by the unknown substance, and since we also know the temperature parameters and the mass, we can solve for it's specific heat capacity.

hayhayz · Answer

delta t for water = 29.5 - 24.0 = 5.5 C 
heat gained by water = 4.18 J/gC x 100.0 g x 5.5 C = 2300 J 
sp ht = J/m dt = 2300 J / (35.5 g x 73.5) = 0.89 J/gC 
Is this right

aaronq · Answer

yep!

aaronq · Answer

good job

aaronq · Answer

wait you didn't write that, you got it from here https://ca.answers.yahoo.com/question/index?qid=20120305055110AAwUvk2

aaronq · Answer

doesn't matter, try to do it yourself though. otherwise you're not going to learn anything

hayhayz · Answer

I wrot on my own paper used the lesson from my school website? The question was actually in my lesson

aaronq · Answer

idk that solution you posted was straight from yahoo answers, letter by letter

hayhayz · Answer

Just a slightly different number was involved and i did this before on an assignment do i need to start sending u proof xD