Find the energy required to melt the copper once it's reached the melting point.
1 answer:
The answer has to depend on the mass of the copper that you're trying to melt.
But there is a formula which you can calculate the energy required if you have the information of the mass. This formula requires the value called specific latent heat of fusion, which is the energy required to melt or freeze a specific mass of copper without changing its temperature but change its state.
E is the total energy required, m is the mass, and lv is the specific latent heat of fusion of copper.
Usually, in questions, lv is given. But we can also look it up online which is around 205kJ kg^-1
this means, to melt or freeze a copper, we need to work the mass multiplied by 205kJ in order to calculate its total energy needed or released.
But there is a formula which you can calculate the energy required if you have the information of the mass. This formula requires the value called specific latent heat of fusion, which is the energy required to melt or freeze a specific mass of copper without changing its temperature but change its state.
E is the total energy required, m is the mass, and lv is the specific latent heat of fusion of copper.
Usually, in questions, lv is given. But we can also look it up online which is around 205kJ kg^-1
this means, to melt or freeze a copper, we need to work the mass multiplied by 205kJ in order to calculate its total energy needed or released.
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Answer: 1. 3.23 m
2. 32.4
3. B adding solvent and C removing solute
Explanation:
1. Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent
where,
n = moles of solute
= weight of solvent
Now put all the given values in the formula of molality, we get
2. Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Therefore, the moles of is 32.4
3. Molarity can be decreased by decreasing the moles of solute and by increasing the volume of solution.
Thus adding solvent and removing solute will decrease the molarity.