Explanation:
a) The amount of heat released by coffee will be absorbed by aluminium spoon.
Thus, 
To calculate the amount of heat released or absorbed, we use the equation:

Also,
..........(1)
where,
q = heat absorbed or released
= mass of aluminium = 45 g
= mass of coffee = 180 g
= final temperature = ?
= temperature of aluminium = 
= temperature of coffee = 
= specific heat of aluminium = 
= specific heat of coffee= 
Putting all the values in equation 1, we get:
![45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]](https://tex.z-dn.net/?f=45%20g%5Ctimes%200.80J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-24%5EoC%29%3D-%5B180%20g%5Ctimes%204.186J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-83%5EoC%29%5D)

80.30 °C is the final temperature.
b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.
So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.
<span>I would say only if one of your data points is the origin. But your experiment could have started with a non-zero velocity, for instance, which would rule out the origin as one of your data points. Even so, a "best fit" is not meant to be perfect, it is only meant to be the best that you can do with your particular data set.</span>
Answer:
Spices
Explanation:
Herbs are the leaves.
Vegetables are the seeds.
Fruits are the seeds.
So spices are the only option left.
<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm
<u>Explanation:</u>
To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

or,

where,
= osmotic pressure of the solution
i = Van't hoff factor = 1 (for non-electrolytes)
= mass of sucrose = 3.40 g
= molar mass of sucrose = 342.3 g/mol
= Volume of solution = 1 L
R = Gas constant = 
T = temperature of the solution = ![20^oC=[20+273]K=293K](https://tex.z-dn.net/?f=20%5EoC%3D%5B20%2B273%5DK%3D293K)
Putting values in above equation, we get:

Hence, the pressure that must be applied to the apparatus is 0.239 atm