Answer:
Removing Energy: Removing energy will cause the particles in a liquid to begin locking into place. A. Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas.
.
Answer:
The heat capacity for the second process is 15 J/K.
Explanation:
Given that,
Work = 100 J
Change temperature = 5 k
For adiabatic process,
The heat energy always same.


We need to calculate the number of moles and specific heat
Using formula of heat


Put the value into the formula


We need to calculate the heat
Using formula of heat

Put the value into the formula


We need to calculate the heat capacity for the second process
Using formula of heat

Put the value into the formula



Hence, The heat capacity for the second process is 15 J/K.
First I will parallel two of the resistors, creating a net 1 ohm. Then I will series that with the remaining 2-ohm resistor, resulting in 3 ohms.