What is the total thermal energy at room temperature 293 K i

What is the total thermal energy at room temperature (293 K) in a gram of (a) lead, (b) dry air (78% N_2, 21% 02, 1% Ar)?

Solution

Total thermal energy = mcdt

where m= mass of the substance

c= specific heat of the substance

dt= temperature difference

a) Lead:

Molecular Mass of lead= 207.2 grams/mol

Now,

Number of moles = Mass /Molecular Mass

= 1 / 207.2

= 0.00482 moles

As MM_{Pb =} 207.2 gm /mol

Now, we have

1 mole = 207.2 grams

0.00482 moles is equal to

0.998704 grams = 1 grams (approx)

Q = mc dt

dt = T₂ - T₁

Here consider T_{1 =} 273 k ( 0 degree celsius)

dt = 293-273 = 20

c_{Pb = 0.128 J/gm K}

_{Q= mc dt}

   _{= 1 x 0.128 x 20 = 2.55 J}

_{b ) Air}

_{C Ar = 5.2 J/g K}

_{Q= 1 x 5.2 x 20 = 104 J}