Question

Assume that the iron block is 175 g and that the beaker contains 75 m hot water, or 75 gince the density of water is 1.00 if you heat the metal directly on the burner to about 275°C and then place it in the room temperature (25°C) water, the temperature of the combined water and iron equilibrates at 75°C Note that the specific heat of water is 4.182, J/g°C Using these values, what is the specific heat of iron in units of J/g•C?

Answer 1

The specific heat of iron is found using the heat exchange formula q = mcΔT, setting the heat lost by the iron equal to the heat gained by the water and solving for the unknown specific heat. After calculation, the specific heat of iron is approximately 0.449 J/g°C.

Step-by-step explanation:

To find the specific heat of iron, we use the principle of the conservation of energy, which states that the heat lost by the iron will be equal to the heat gained by the water. The formula used is q = mcΔT, where q is the heat exchanged, m is the mass, c is the specific heat, and ΔT is the change in temperature.

• Let qw be the heat gained by the water and qFe be the heat lost by the iron.
• Because qw = -qFe, we have mwcwΔTw = -mFecFeΔTFe.
• Substitute the given values: (75 g)(4.182 J/g°C)(75°C - 25°C) = -(175 g)(cFe)(275°C - 75°C)
• Calculate the left side to find the heat absorbed by the water: qw = 15,682.5 J.
• Now solve for the specific heat of iron, cFe: 15,682.5 J = (175 g)(cFe)(200°C), hence cFe = 15,682.5 J / (175 g × 200°C).
• After calculating, the specific heat of iron is found to be approximately 0.449 J/g°C.