I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.
In fact ! You didn't even say which frame the '12 km/hr' of his
bike is measured in.
So there are several different ways this could go. I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.
-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.
-- He throws a ball at Keesha, at 5 km/hr relative to his own face.
-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.
Answer:
The minimum work per unit heat transfer will be 0.15.
Explanation:
We know the for a heat pump the coefficient of performance (
) is given by
![C_{HP} = \dfrac{Q_{H}}{W_{in}}](https://tex.z-dn.net/?f=C_%7BHP%7D%20%3D%20%5Cdfrac%7BQ_%7BH%7D%7D%7BW_%7Bin%7D%7D)
where,
is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature
and
is the required input and is given by
,
being magnitude of heat transfer between cyclic device and low-temperature
. Therefore, from above equation we can write,
![&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}](https://tex.z-dn.net/?f=%26%26%20%5Cdfrac%7BQ_%7BH%7D%7D%7BW_%7Bin%7D%7D%20%3D%20%5Cdfrac%7BQ_%7BH%7D%7D%7BQ_%7BH%7D%20-%20Q_%7BL%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B1%20-%20%5Cdfrac%7BQ_%7BL%7D%7D%7BQ_%7BH%7D%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B1%20-%20%5Cdfrac%7BT_%7BL%7D%7D%7BT_%7BH%7D%7D%7D)
Given,
and
. So, the minimum work per unit heat transfer is given by
![\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15](https://tex.z-dn.net/?f=%5Cdfrac%7BW_%7Bin%7D%7D%7BQ_%7BH%7D%7D%20%3D%20%5Cdfrac%7BT_%7BH%7D%20-%20T_%7BL%7D%7D%7BT_%7BH%7D%7D%20%3D%20%5Cdfrac%7B540%20-%20460%7D%7B540%7D%20%3D%200.15)
Answers:
1A) Al203
1B) SF6
2) Fe203 - iron oxide
I’m assuming that’s m^3? If so then simply divide 160,000 by 20 and you get the answer.
8,000 kg/m^3