For each engine calculate δe = qh−wout−qc, where qh is the amount of heat transferred from the hot reservoir, qc is the amount of heat transferred to the cold reservoir, wout is the energy output of the heat engine. qh, wout, qc are positive quantities.

The given equation is an expression of the first law of thermodynamics,

that states that the energy in a closed system is conserved.

The values for each engine are;

(a) ΔE = 0 J

(b) ΔE = -1 J

(c) ΔE = 0 J

Reasons:

Given equation; ΔE = - -

Required:

To calculate ΔE = - - for the three heat engines;

Solution:

(a) = 50 J, = 30 J, = 20 J

Therefore;

ΔE = 50 J - 30 J - 20 J = 0 J

(b) = 10 J, = 4 J, = 7 J

Therefore;

ΔE = 10 J - 4 J - 7 J = -1 J

(b) = 30 J, = 10 J, = 20 J

Therefore;

ΔE = 30 J - 10 J - 20 J = 0 J

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The energy output of the of the heat engine at various location is  0J, -1J and 0J

Explanation:

The change in internal energy of the process is according to the diagram

          = 0 J

            = -1 J

          = 0 J

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.