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·°C, what is the heat capacity of the copper in J/°C?

Question

Quizplus · Accepted Answer

The amount of heat lost by copper = the amount of heat gained by water

q lost = q gained

m_copper * c_copper * (T_final - T_initial) = m_water * c_water * (T_final - T_initial)

Since the final temperature is the same for both, we can simplify to:

m_copper * c_copper * (T_final - T_initial) = m_water * c_water * (T_final - T_initial)

c_copper = m_water * c_water * (T_final - T_initial) / (m_copper * (T_final - T_initial))

c_copper = m_water * c_water / m_copper

c_copper = (50.0 g) * (4.18 J/g·°C) / (mass of copper)

mass of copper = initial temperature difference / (final temperature difference / (mass of copper * specific heat))

mass of copper = (100.0°C - 25.0°C) / (25.0°C - 20.0°C) / (50.0 g * 4.18 J/g·°C)

mass of copper = 0.012 kg

c_copper = (50.0 g) * (4.18 J/g·°C) / (0.012 kg)

c_copper = 173.3 J/°C

Therefore, the heat capacity of copper is 13.9 J/°C (since heat capacity = mass * specific heat).

A Piece of Copper Metal Is Initially at 100