Answer:
Explanation:
Hello,
In this case, considering the given diameter which is related to a radius of 5.0 mm and the formula for the calculation of the volume of the sphere, its volume in cubic centimeters (5.00 mm = 0.5 cm) is then:
In such a way, the density turns out:
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Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
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Answer:
There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C
Explanation:
To decrease the temperature of the solution there are necessaries:
4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y
8368J + 83,68J/gX = Y <em>(1)</em>
Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.
Also, the energy Y will be:
Y = 25700J/mol×X
Y = 321J/g X <em>(2)</em>
Replacing (2) in (1)
8368J + 83,68J/g X = 321J/g X
8363J = 237,32J/gX
<em>X = 35,2g</em>
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Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C
I hope it helps!