Question

A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the calorimeter is 31.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

Answer 1

Answer:

S(metal) = 0.66J/g°C

Explanation:

We can find specific heat of a material, S, using the equation:

q = m*S*ΔT

Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.

The heat given by the metal is equal to the heat that water absorbs, that is:

m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)

Where:

m(Metal) = 76.0g

S(metal) = ?

ΔT(Metal) = 96.0°C-31.0°C = 65.0°C

m(Water) = 120.0g

S(water) = 4.184J/g°C

ΔT(water) = 31.0°C-24.5°C = 6.5°C

Replacing:

76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C

S(metal) = 0.66J/g°C

The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.