Question

11. A bronze statue that weighs 5.80 kilograms is heated to a uniform temperature of 81.0 degrees Celsius. The statue is then placed in a bath containing 15.0 kilograms of water at 4.0 degrees _ IT - 8 at jus. f the specific heat of bronze is 0 435 /8°C, what wil be the equilibrium terperature of the water bath, assuming no heat is lost? ( Remember Q = mcAT,

Answer 1

To determine the specific heat capacity of the metal, we can use the equation Q = mcΔT, where Q represents the heat energy transferred, m represents the mass of the substance, c represents the specific heat capacity, and ΔT represents the change in temperature. By setting the heat energy for the metal and water equal, we can solve for the specific heat capacity of the metal. In this case, the specific heat capacity of the metal is approximately 0.879 J/g°C.

Step-by-step explanation:

To answer this question, we can use the equation Q = mcΔT, where Q represents the heat energy transferred, m represents the mass of the substance, c represents the specific heat capacity, and ΔT represents the change in temperature.

For the metal, we have m = 34.5 g, ΔT = (39°C - 75°C) = -36°C, and c is what we need to find.

For the water, we have m = 64.0 g, ΔT = (39°C - 25°C) = 14°C, and c = 4.18 J/g°C.

Setting the heat energy for the metal and water equal, we can solve for c:

mc(ΔT) for metal = mc(ΔT) for water

(34.5 g)(c)(-36°C) = (64.0 g)(4.18 J/g°C)(14°C)

Solving for c, we find c ≈ 0.879 J/g°C. Therefore, the specific heat capacity of the metal is approximately 0.879 J/g°C.