Answer:
Explanation:
As we know that charge is always conserved
so here we have
initial total charge = final total charge
also we know that gamma rays are chargeless and massless particles
so we have
charge on deuterium is given as
Q = charge on proton + charge on neutron - charge on gamma
so we have
Answer:
a)ΔS₁ = - 9.9 J/K
ΔS₂ = 69 J/K
b)The entropy change for the rod = 0 J/K
c)ΔS = 59.1 J/K
Explanation:
Given that
T₁ = 699 K
T₂= 101 K
Q= 6970 J
Change in entropy given as
For 699 K:
ΔS₁ = - 9.9 J/K ( Negative because heat is leaving from the system)
For 101 K;
ΔS₂ = 69 J/K
The entropy change for the rod = 0 J/K
Entropy change for the system
ΔS = ΔS₂ + ΔS₁
ΔS = 69 -9.9 J/K
ΔS = 59.1 J/K