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What is the heat capacity of a 31 gram substance
that releases -3452J of heat energy and has an initial
temperature of 100°C and a final temperature of 45.1°C?

Respuesta :

Answer:

2 J/g Celcius

Explanation:

Find delta T: final temp - initial temp

q=energy, m= mass, c= specific heat. delta T= change in temp

45.1 C-100 C= -54.9 C

Rearrange energy equation to solve of c (specific heat):

q=m*c*delta T   --> c=q/(m*delta T)

Now, just plug in the numbers:

c= (-3452J)/(31g*-54.9 C)

... to get

c=2J/(g C) (rounded to 1 sig fig due to the 100 stated in the original problem)

Hope that helps!

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