Answer
Given,
Energy absorbed, ![Q_H = 1.69\ kJ](https://tex.z-dn.net/?f=Q_H%20%3D%201.69%5C%20kJ)
Energy expels,![Q_C = 1.25\ kJ](https://tex.z-dn.net/?f=%20Q_C%20%3D%20%201.25%5C%20kJ)
Temperature of cold reservoir, T = 27°C
a) Efficiency of engine
![\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cdfrac%7BQ_H%20-%20Q_C%7D%7BQ_H%7D%5Ctimes%20100)
![\eta = \dfrac{1.69 - 1.25}{1.69}\times 100](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cdfrac%7B1.69%20-%201.25%7D%7B1.69%7D%5Ctimes%20100)
![\eta =26.03 %](https://tex.z-dn.net/?f=%5Ceta%20%3D26.03%20%25)
b) Work done by the engine
![W = Q_H- Q_C](https://tex.z-dn.net/?f=W%20%3D%20Q_H-%20Q_C)
![W =1.69 - 1.25](https://tex.z-dn.net/?f=W%20%3D1.69%20-%201.25)
![W = 0.44\ kJ](https://tex.z-dn.net/?f=%20W%20%3D%200.44%5C%20kJ)
c) Power output
t = 0.296 s
![P = \dfrac{W}{t}](https://tex.z-dn.net/?f=P%20%3D%20%5Cdfrac%7BW%7D%7Bt%7D)
![P = \dfrac{0.44}{0.296}](https://tex.z-dn.net/?f=P%20%3D%20%5Cdfrac%7B0.44%7D%7B0.296%7D)
![P = 1.486\ kW](https://tex.z-dn.net/?f=%20P%20%3D%201.486%5C%20kW)
Given that the density of heptane is
![d_h=\frac{0.684g}{mL}](https://tex.z-dn.net/?f=d_h%3D%5Cfrac%7B0.684g%7D%7BmL%7D)
The mass of heptane is
![m_h=31\text{ g}](https://tex.z-dn.net/?f=m_h%3D31%5Ctext%7B%20g%7D)
The density of water is
![d_w=\frac{1g}{mL}](https://tex.z-dn.net/?f=d_w%3D%5Cfrac%7B1g%7D%7BmL%7D)
The mass of water is
![m_w=37\text{ g}](https://tex.z-dn.net/?f=m_w%3D37%5Ctext%7B%20g%7D)
The volume of heptane will be
![\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V_h%3D%5Cfrac%7Bm_h%7D%7Bd_h%7D%20%5C%5C%20%3D%5Cfrac%7B31%7D%7B0.684%7D%20%5C%5C%20%3D45.32%5Ctext%7B%20mL%7D%20%5Cend%7Bgathered%7D)
The volume of water will be
![\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V_w%3D%5Cfrac%7Bm_w%7D%7Bd_w%7D%20%5C%5C%20%3D%5Cfrac%7B37%7D%7B1%7D%20%5C%5C%20%3D37%5Ctext%7B%20mL%7D%20%5Cend%7Bgathered%7D)
Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.
The total volume of liquid in the cylinder will be
![\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3DV_h%2BV_w%20%5C%5C%20%3D45.32%2B37%20%5C%5C%20%3D82.32%5Ctext%7B%20mL%7D%20%5Cend%7Bgathered%7D)
The total volume of liquid in the cylinder will be 82.32 mL.
Answer: They create calcuim chloride, CaCl2
Answer: I'm not sure what it needs to be rounded to, but I got 37.53501401 m/s
Explanation: The formula for speed is speed = distance/time. You plug in the distance (13.40) and the time (0.357), then divide 13.40 by 0.357
I hope this helps! :)
I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.
Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .
Here's the formula for the distance an object falls from rest
in a certain time:
Distance = (1/2) (gravity) (time)²
On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...
11.2 cm = (1/2) (9.8 m/s²) (time)²
or
0.112 meter = (4.9 m/s²) (time)²
Divide each side
by 4.9 m/s² : (0.112 m) / (4.9 m/s²) = time²
(0.112 / 4.9) sec² = time²
Square root
each side: time = √(0.112/4.9 sec²)
= √ 0.5488 sec²
= 0.74 second (rounded)