To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

Where,
Q = Heat
U = Internal Energy
By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore
Work is done ON the system
Heat is extracted FROM the system
Therefore the value of the Work done on the system is -158.0J
Answer:
45200J
Explanation:
Given parameters:
Heat of vaporization of water = 2260J/g
Mass of steam = 20g
Temperature = 100°C
Unknown:
Energy released during the condensation = ?
Solution:
This change is a phase change and there is no change in temperature
To find the amount of heat released;
H = mL
m is the mass
L is the latent heat of vaporization
Insert the parameters and solve;
H = 20g x 2260J/g
H = 45200J
Answer:
The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0
An since we can't divide a number by 0, the question is wrong
Answer:
the extension would be less the new extension might be 3 cm
Explanation: