Complete Question
A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of 650 K and 270 K and performs 4.3 kJ of net work and rejects 8.00 kJ of heat in a single cycle. The thermal efficiency of this heat engine is closest to A) 0.35 B) 0.31. C) 0.28. D) 0.38. E) 0.42.
Answer:
The correct option is A
Explanation:
From the question we are told that
The first operating temperature is ![T_1 = 650 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%20650%20%5C%20K)
The second operating temperature is ![T_2 = 270 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20270%20%5C%20K)
The net workdone is
( output of the engine )
The amount of heat energy rejected is ![H = 8.00 \ kJ = 8.00 *10^{3 } \ J](https://tex.z-dn.net/?f=H%20%20%3D%20%208.00%20%5C%20kJ%20%20%3D%20%208.00%20%2A10%5E%7B3%20%7D%20%5C%20%20J)
Generally a heat engine convert heat from a high temperature to mechanical energy and then reject the remaining heat so the absorbed by the engine is
![W + H](https://tex.z-dn.net/?f=W%20%2B%20H)
Generally the thermal efficiency is mathematically represented as
![\eta = \frac{out}{In} * 100](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%20%20%5Cfrac%7Bout%7D%7BIn%7D%20%20%2A%20100)
Here out is the output of the engine
and in is the input of the engine
![\eta = \frac{W}{W + H} * 100](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%20%20%5Cfrac%7BW%7D%7BW%20%2B%20H%7D%20%20%2A%20100)
=> ![\eta = \frac{4.3}{4.3 + 8} * 100](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%20%20%5Cfrac%7B4.3%7D%7B4.3%20%2B%208%7D%20%20%2A%20100)
=> ![\eta = 0.35](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%200.35)
Explanation:
weight=mass ×acceleration
weight=25×9.8
=245
The ice melts a liquid the molecules spread out and when a liquid changes to a gas the molecules spread out more.
Answer:
The mass of the body is 2.2 kg.
Explanation:
Given that,
Force acting on a body 1, ![F_1=20\ N](https://tex.z-dn.net/?f=F_1%3D20%5C%20N)
Force acting on a body 2, ![F_2=35\ N](https://tex.z-dn.net/?f=F_2%3D35%5C%20N)
Direction, ![\theta=80^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D80%5E%7B%5Ccirc%7D)
Acceleration, ![a=20\ m/s^2](https://tex.z-dn.net/?f=a%3D20%5C%20m%2Fs%5E2)
To find,
Mass of the body.
Solution,
Let F is the net force acting on the body. It is given by :
![F=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}](https://tex.z-dn.net/?f=F%3D%5Csqrt%7BF_1%5E2%2BF_2%5E2%2B2F_1F_2%5C%20cos%5Ctheta%7D)
![F=\sqrt{20^2+35^2+2\times 20\times 35\ cos(80)}](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B20%5E2%2B35%5E2%2B2%5Ctimes%2020%5Ctimes%2035%5C%20cos%2880%29%7D)
F = 43.22 N
Second law of motion is given by :
F = ma
![m=\dfrac{F}{a}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BF%7D%7Ba%7D)
![m=\dfrac{43.22\ N}{20\ m/s^2}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B43.22%5C%20N%7D%7B20%5C%20m%2Fs%5E2%7D)
m = 2.16 kg
or
m = 2.2 kg
So, the mass of the body is 2.2 kg.