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?

The quantity of heat lost when the piece of aluminum cools from 415°C to 22°C is -8501.99 Joules.

Given the following data:

To find how much heat is lost when the piece of aluminum cools:

Mathematically, heat capacity or quantity of heat is given by the formula;

Where:

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Substituting the values into the formula, we have:

Q = -8501.99 Joules.

Read more: link

Q = - 8501.99 j

Explanation:

Given data:

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

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole