Question

A 1.30 × 10² g piece of metal is heated to 285 °C and dropped into 81.0 g of water at 26.0 °C. If the final temperature of the water and metal is 58.5 °C, what is the specific heat of the metal?

a) 0.25 J/(g°C)
b) 0.50 J/(g°C)
c) 0.75 J/(g°C)
d) 1.00 J/(g°C)

Answer 1

To determine the specific heat of the metal, we use the conservation of energy principle, which sets the heat lost by the metal equal to the heat gained by water. By solving the resulting equation, we can find the specific heat value of the metal.

Step-by-step explanation:

To find the specific heat of the metal, we can use the principle of conservation of energy, which implies that the heat lost by the hot metal will equal the heat gained by the cooler water. The formula we use is 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.

The formula for the heat gained by water is: q(water) = m(water) × c(water) × ΔT(water)

The formula for the heat lost by the metal is: q(metal) = m(metal) × c(metal) × ΔT(metal)

Since the heat lost by metal equals the heat gained by water, we have m(metal) × c(metal) × ΔT(metal) = m(water) × c(water) × ΔT(water).

Inserting the values from the question, we get:

1.30 × 10² g × c(metal) × (58.5 °C - 285 °C) = 81.0 g × 4.184 J/g°C × (58.5 °C - 26.0 °C)

Solving for c(metal), we find:

c(metal) = (81.0 g × 4.184 J/g°C × 32.5 °C) / (1.30 × 10² g × 226.5 °C)

Calculating yields the specific heat of the metal.