A piece of metal of mass 19 g at 96◦C is placed in a calorimeter containing 53.2 g of water?
Update:
A piece of metal of mass 19 g at 96◦C is placed
in a calorimeter containing 53.2 g of water at
25◦C. The final temperature of the mixture
is 40.6
◦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
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Answers & Comments
The heat energy that is absorbed by the water is equal to the heat energy that released by the metal.
Q = mass * specific heat * ∆ T
For water, the specific heat is 4186 J/(kg * ˚C)
Q = 0.0532 * 4186 * (40.6 – 25) = 3774.04212 J
Q = 0.019 * x * (96 – 40.6) = x * 1.0526
x * 1.0526 = 3774.04212
x = 3774.04212 ÷ 1.0526
This is approximately 3300 J/(kg * ˚C).
How am I supposed to know if it was?