How much energy (in J) is lost when a sample of iron with a mass of 28.3 g cools from 66.0 degrees celsius to 24.0 degrees celsius.

-533.68J

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat required (Joules)

m = mass of substance (g)

c = specific heat of substance (J/g°C)

∆T = change in temperature (°C)

According to the information provided in this question;

Q = ?

m = 28.3 g

c of iron = 0.449 J/g°C

∆T = 24°C - 66°C = -42°C

Using the formula, Q = m × c × ∆T

Q = 28.3 × 0.449 × (-42)

Q = -533.68

Q = -533.68J

This means that 533.68J is lost to the environment when iron of 28.3g cools from 66°C to 24°C.