Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).
Answer:
a
b
Explanation:
Generally the force of attraction between this two irons is mathematically represented as
![F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}](https://tex.z-dn.net/?f=F%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20%5BQ_%7BLi%7D%20%20%5D%20%2A%20%5BQ_%7BO%7D%20%20%5D%20%20%7D%7B%20r%5E2%7D)
Here k is known as the proportionality constant with value 
substituting -2 for 
i.e the charge on oxygen , +1 for 
i.e the charge on Lithium and ![[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}](https://tex.z-dn.net/?f=%20%5B0.140%20%2B%200.068%20%5D%20nm%3D%200.208%20nm%20%3D%20%200.208%2A10%5E%7B-9%7D%20%20%20)
for r
So
Generally the force of repulsion will be the magnitude but different direction to the force o attraction
So Force of repulsionn is
2. kinetic energy: due to it being transferred through collisions
•responsiveness to the environment;
•growth and change;
•ability to reproduce;
•have a metabolism and breathe;
•maintain homeostasis;
•being made of cells; and.
•passing traits onto offspring.
Explanation: Solution
1.
Gravitational potential energy
U=mgh=500*9.8*50
U=245000 J
2.
Kinetic energy is present at bottom of the hill
K=(1/2)mV2=(1/2)*500*27.82
K=193210 J
3.
Work done by friction
W=193210-245000=-51790 J
