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). (2 points)

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate things. We do as follows:

Heat gained = Heat lost
mC(T2-T1) = - mC(T2-T1) 
31.5C (102.4 - 32.5) = 103.5(4.18)(32.5 - 24.5)
C = 1.57 J/C-g

Hope this helps.