Question

95.0 g of an unknown metal at 100°C is placed in 800 g of water at 25.0°C. The final temperature of the water is 29.8°C. What was the specific heat of the metal?

A. 0.5 J/g°C

B. 1.0 J/g°C

C. 2.0 J/g°C

D. 4.0 J/g°C

Answer 1

To find the specific heat capacity of the unknown metal, the heat lost by the metal is set equal to the heat gained by the water. The specific heat capacity is calculated with the formula q = mcΔT, using the given masses and temperatures. A calculation error may have occurred as the result does not match the provided options.

Step-by-step explanation:

To determine the specific heat of the unknown metal, we must apply the concept of heat transfer, where the heat lost by the hot metal will be equal to the heat gained by the colder water. We can use the formula: q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the metal: q_metal = (95.0 g)(c)(100°C - 29.8°C)

For the water: q_water = (800 g)(4.18 J/g°C)(29.8°C - 25.0°C)

Since the metal loses heat and the water gains heat: -q_metal = q_water.

Now we calculate the heat gained by the water: q_water = (800 g)(4.18 J/g°C)(4.8°C) = 16054.4 J.

That implies: -(95.0 g)(c)(70.2°C) = 16054.4 J. Solving for c gives us a specific heat capacity for the metal: c = 16054.4 J / (95.0 g)(70.2°C), which after calculation gives c = 2.4 J/g°C. However, this value is not an option provided in the question, implying either a mistake in the setup or in the calculations. We would need to recheck our work.