Answer:
The temperature of the core raises by
every second.
Explanation:
Since the average specific heat of the reactor core is 0.3349 kJ/kgC
It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius
Thus reactor core whose mass is
will require

energy to raise it's temperature by 1 degree Celsius in 1 second
Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by
The total distance traveled by the body is the sum of the distance it traveled for the first five seconds which is 10 m and that of the next five seconds which is 30 meters. Thus, the total distance traveled is 40 m. Dividing this by the time, will give the average speed. The average speed is therefore, 4 m/s. The answer is letter C.
Ratatouille is my momCan you lend me 700 because I used my toaster as a bath heater and now my legs are gone plz I need money for bandages
I would say C. if im wrong, im sorry. but if im right your welcome. i didnt guess tho.
Answer:
Heat transfer during the process = 0
Work done during the process = - 371.87 KJ
Explanation:
Initial pressure
= 0.02 bar
Initial temperature
= 200 K
Final pressure
= 0.14 bar
Gas constant for helium R = 2.077 
This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,
= ![[\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BP_%7B2%7D%20%7D%7BP_%7B1%7D%20%7D%20%5D%5E%7B%5Cfrac%7B%5Cgamma%20-%201%7D%7B%5Cgamma%7D%20%7D)
Put all the values in above formula we get,
⇒
= ![[\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7B0.14%20%7D%7B0.02%20%7D%20%5D%5E%7B%5Cfrac%7B1.4%20-%201%7D%7B1.4%7D%20%7D)
⇒
= 1.74
⇒
= 348.72 K
This is the final temperature of helium.
For isentropic polytropic process heat transfer to the system is zero.
⇒ ΔQ = 0
Work done W = m × (
-
) × 
⇒ W = 1 × ( 200 - 348.72 ) × 
⇒ W = 371.87 KJ
This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.