Question

31.8 grams of an unknown substance is heated to 55.0 degrees Celsius and then placed into a calorimeter containing 60.0 grams of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 12.4 degrees Celsius, what is the specific heat of the unknown substance?

Answer 1

question incomplete as in initial temp of unknown substance not mentioned

Answer 2

The question doesn't make sense.
Since the unknown substance is hotter, when placing in a cooler place, the final temp of water should remain unchanged/ rise instead of decrease. In this case, it decreases. Can you check the question again?

Answer 3

And for doing this type of question, you can do it in this way
Assume there is no heat loss to the surrounding
By conservation of energy
Heat loss = Heat gained
Then, calculate energy loss/gain by using E= mcΔT
That is
31.8 x (specific heat capacity of unknown substance) x (temperature change of water)
= 60 x (specific heat capacity of water) x (temperature change of water)