Question

125.0 g of an unknown substance is heated to 97.0°C. It is then placed in a calorimeter than contains 250g of water with an initial temperature of 20.0°C. The final temperature reached by the calorimeter is 23.5°C. What is the specific heat of the unknown substance? The specific heat of water is 4.18 J/ (°C × g)

Answer 1

answer choices:
0.285 J/ (°C × g)
0.398 J/ (°C × g)
0.729 J/ (°C × g)
1.24 J/ (°C × g)

Answer 2

formula:
Q (energy) = (mass)(specific heat)(change in temp)

You have two equations, and Q is equal. So:
(mass of water)(specific heat of water)(change in temp of water)=(mass of substance)(specific heat of substance)(change in temp of substance).

(250 g)(4.18 J/(K x g))(3.5 K)=(125 g)(C)(73.5 K)

solving for C to equal 0.398 J/ (°C × g).

Answer 3

are you taking a test and cheating? because if you are, I have to stop.

Answer 4

Well it's a study guide for my final.