A 1.00 g piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20.0°C. After stirring, the final temperature of both copper and water is 25.0°C. Assuming no heat losses, and that the specific heat (capacity) of water is 4.18 J/(g·K), what is the heat capacity of the copper in J/K?

Question

Quizplus · Accepted Answer

Firstly, we need to find the amount of heat lost by the copper and the amount of heat gained by the water to reach the final temperature. We can use the formula:

q = m·c·ΔT

where
q = heat energy (in J)
m = mass (in g)
c = specific heat capacity (in J/(g·K))
ΔT = change in temperature (in K)

For the water:
q = (50.0 g)·(4.18 J/(g·K))·(25.0°C - 20.0°C)
q = 1045 J

For the copper:
q = -(1.00 g)·c·(100.0°C - 25.0°C)
q = -(75 g·K)·c

Since the heat lost by the copper is equal to the heat gained by the water, we can set the two equations equal to each other:

1045 J = -(75 g·K)·c

Solving for c, we get:

c = -1045 J / (75 g·K)
c = -13.9 J/(g·K)

The negative sign just means that the copper lost heat while the water gained heat, which is expected. We take the absolute value of c, which is 13.9 J/(g·K). Therefore, the heat capacity of copper is 13.9 J/(g·K).

A 100 G Piece of Copper Metal Is Initially at 100