Aluminum has a specific heat of 0.902 j/gC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a temperature of 415 degrees to 22 degrees?

Q = - 8501.99 j

Explanation:

Given dа ta:

Specific heat of Al = 0.902 j/g.°C

Heat lost = ?

Mass of sample = 23.984 g

Initial temperature = 415°C

Final temperature = 22°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 22°C - 415°C

ΔT = - 393°C

Q = m. c. ΔT

Q = 23.984 g * 0.902 j/g.°C * - 393°C

Q = - 8501.99 j