Question

A 55.0 g piece of metal is heated in boiling water to 99.8 ⁰C and then dropped into cool water in an isolated beaker. The beaker contains 225.0 g of water and its temperature (before the metal was dropped in) was 21.0 ⁰C . The final temperature of the metal and water is 23.1 ⁰C( thermal equilibrium). What is the specific heat capacity of the metal? Cp of water is 4.184 J/g⁰C

Answer 1

The specific heat capacity of the metal, when dropped into water and reaching thermal equilibrium, is calculated using the conservation of energy. After equating the heat lost by the metal to the heat gained by the water and solving the equation, it is found to be approximately 0.469 J/g°C.

Step-by-step explanation:

To find the specific heat capacity of the metal, we can use the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water. We set the heat lost by the metal equal to the heat gained by the water and solve for the specific heat capacity of the metal (Cp_metal).

The equation for the heat lost or gained is Q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. Since the system reaches thermal equilibrium, we have:

• Heat lost by metal: Q_metal = (mass_metal)(Cp_metal)(change in temp_metal)
• Heat gained by water: Q_water = (mass_water)(Cp_water)(change in temp_water)

Q_metal = -Q_water (the negative sign indicates heat lost by metal and gained by water)

(55.0 g)(Cp_metal)(99.8 °C - 23.1 °C) = -(225.0 g)(4.184 J/g°C)(23.1 °C - 21.0 °C)

Now solve for Cp_metal:

Cp_metal = -(225.0 g)(4.184 J/g°C)(2.1 °C) / (55.0 g)(76.7 °C)

Cp_metal = -1981.8 J / 4218.5 g°C

Cp_metal ≈ 0.469 J/g°C

The specific heat capacity of the metal is approximately 0.469 J/g°C.