A 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

The heat lost by the copper is equal to the heat gained by the water. The heat lost by the copper is given by:Q = m * c * ΔTwhere m is the mass of the copper, c is the specific heat of the copper, and ΔT is the change in temperature of the copper.The heat gained by the water is given by:Q = m * c * ΔTwhere m is the mass of the water, c is the specific heat of the water, and ΔT is the change in temperature of the water.Setting these two equations equal to each other, we get:m * c * ΔT = m * c * ΔTSolving for c, we get:c = m * ΔT / m * ΔTPlugging in the values given in the problem, we get:c = (50.0 g) * (25.0°C - 100.0°C) / (100.0 g) * (25.0°C - 20.0°C)c = 2.79 J/K

A Piece of Copper Metal Is Initially at 100