Question

A piece of metal of mass 26 g at 117◦C is placed in a calorimeter containing 56 g of water at 25◦C. The final temperature of the mixture is 58.7◦C. What is the specific heat capacity of the metal? Assume that there is no energy lost to the surroundings.
Answer in units of J . g ·◦ C

Answer 1

Step 1) Determine the amount of energy the water absorbed, or gained, using q= mc∆T

q=(56g)(4.184 J/g°c)(58.7-25)

q=+7896.0448 J

Step 2) Let's think about what this means. The water must have gained 7896.0448 J of energy because its energy is positive.

This makes sense too, since the temperature of the water increased, and thus it must have gained energy. Knowing this, we can conclude that the metal placed into the water must have released 7896.0448 J of energy.

Since we know the metal released 7896.0448 J of energy, when we solve for the specific heat capacity of the metal, we have to put this number in as negative.

Step 3) Using q= mc∆T, solve for the specific heat capacity of the metal

-7896.0448 J=(26g)(c)(58.7-117)

c=5.209160047 J/g°c

ANSWER: the specific heat capacity of the metal is 5.2 J/g°c