To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

Here,
Q = Total Heat
T = Temperature
The total change of entropy from a cold object to a hot object is given by the relationship,

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'
Change in entropy
is smaller than 
Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
Answer:
Electric force, 
Explanation:
Given that,
Electric charge 1, 
Electric charge 2, 
Distance, 
To find,
The electric force between these two sets of charges.
Solution,
There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

k is the electrostatic constant


So, the electric force between these two sets of charges is
.