The specific heat of water is 4.18 J/g•°C. How much heat does 225.0 g of water release when it cools from 85.5°C to 50.0°C? Ente
2 answers:
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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!
Answer : q = 6020 J, w = -6020 J, Δe = 0
Solution : Given,
Molar heat of fusion of ice = 6020 J/mole
Number of moles = 1 mole
Pressure = 1 atm
Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.
The relation between heat and molar heat of fusion is,
(in terms of mass)
or, (in terms of moles)
Now we have to calculate the value of q.
When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.
So, the value of
Now we have to calculate the value of w.
Formula used :
where, q is heat required, w is work done and is internal energy.
Now put all the given values in above formula, we get
w = -6020 J
Therefore, q = 6020 J, w = -6020 J, Δe = 0