Question

A cool water sample absorbed 1,710 J of energy from some metal. The temperature of the 59.0 g piece of metal changed from 99.5°C to 27.8°C.

What is the specific heat of the metal? q,H₂O=1,710J
Cₘₑₜₐₗ=[?] J/g·°C

Answer 1

To find the specific heat of the metal, we use the energy absorbed by the water, which is equal to the energy lost by the metal. The calculation is done using the formula for heat transfer, resulting in a specific heat of 0.404 J/g°C for the metal.

Step-by-step explanation:

To calculate the specific heat of a metal using the energy absorbed by water, we use the formula q = m × c × ΔT, where q is the amount of heat in joules, m is the mass in grams, c is the specific heat capacity in J/g°C, and ΔT is the change in temperature in °C.

In this case, the water absorbed 1,710 J of energy.

Therefore, we can assume the metal released the same amount of energy (because energy is conserved, so what the water gained, the metal lost). The formula can be rearranged to solve for the specific heat capacity of the metal (Cmetal):

Cmetal = q / (m × ΔT)

Put the known values into the formula:

Cmetal = 1,710 J / (59.0 g × (99.5°C - 27.8°C))

Cmetal = 1,710 J / (59.0 g × 71.7°C)

Then perform the calculation:

Cmetal = 1,710 J / (4,227.3 g°C)

Cmetal = 0.404 J/g°C

The specific heat of the metal is therefore 0.404 J/g°C.