jocejay83681
08.10.2019 •
Chemistry
How much boiling water at 100 ∘c must you add to this beaker so that the final temperature of the mixture will be 77 ∘c?
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Ответ:
Now, you should know that there are two types of heat energy: latent heat and sensible heat. Latent heat is the heat you add or remove at a constant temperature when there is a phase change occurring. On the other hand, sensible heat is the heat you add or remove as you raise or reduce the temperature of a system. For a boiling water, a phase change occurs, so we're going to use latent heat for that one. The heat used to raise the temperature to 77°C is the sensible heat.
The equation for latent hear is mΔH, where ΔH for phase change from liquid to gas is called latent heat of vaporization. For water, ΔH = 334 Joules/gram. The equation for sensible heat is mCpΔT, where Cp is the specific heat of a substance. For water, Cp = 4.186 J/g-°C.
However, we need to know what's inside the beaker initially. But there is no given data for this. Let's just assume that there is 100 grams of water at 25°C in the beaker initially. Therefore, our working equation would be:
mCpΔT = mΔH
We equate them because, as we said, heat is just transferred between the two. So, the energy is just equal. Substituting the values:
(10 g)(4.186 J/g-°C)(77°C - 25°C) = m(334 J/g)
Solving for m,
m = 6.517 g
Therefore, if there is initially 10 grams at 25°C of water in the beaker, then you would need 6.517 g of boiling water to be able to raise the temperature to 77°C.
Ответ:
The appropriate solution is "6.818".
Explanation:
The given value is:
Concentration of HCl (Actual value),
= 0.33 M
Concentration of HCl (Experimental value),
= 0.3515 M
Now,
The percentage error (%) will be:
=
On substituting the given values, we get
=
=
=
=