Answer:-16.3 kJ
Step-by-step explanation:
Given
Pressure inside Cylinder=52,600 Pa
Piston Moved Slowly inwards 0.15 m
Area of piston(A)
![=(\pi )/(4)d^2](https://img.qammunity.org/2020/formulas/physics/college/en4gvz1arp6l6fz3j5vnrckf1zqjkogvy5.png)
![A=0.2922 m^2](https://img.qammunity.org/2020/formulas/physics/college/fqngjticc7vc9wbf3n7bil2jlzshwo57jd.png)
Change in volume=A\Delta L[/tex]
![\Delta V=0.292* 0.15=0.04384 m^3](https://img.qammunity.org/2020/formulas/physics/college/3q4zyutwkrowc6it3r8a5i0amql9bflt96.png)
Heat removal=18,600 J
From First Law of thermodynamics
![\Delta U=Q-W](https://img.qammunity.org/2020/formulas/physics/middle-school/djrk88m6vhal5oaba5g4kvot72dvkur2ma.png)
![W=P\Delta V](https://img.qammunity.org/2020/formulas/physics/college/lh6n5dqids8sz1ngwbvikkd8u3zabyokiv.png)
![W=52600* 0.04384=-2.3061 kJ](https://img.qammunity.org/2020/formulas/physics/college/q8e515467zrfdo5gagnmhx0g9r6ecwhh6x.png)
as volume is decreasing
Q=-18.6 kJ
![\Delta U=-18.6+2.3061](https://img.qammunity.org/2020/formulas/physics/college/cksviov5mmwyfqfb925vt6bvd20181q6dy.png)
![\Delta U=-16.29 \approx 16.3 kJ](https://img.qammunity.org/2020/formulas/physics/college/a8jfg8c7tsefipylom6jzlum4bmu45xyn0.png)