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2 years ago

Differences between freezing point and melting point (Atleast 5 differences)

Physics

1 answer:

labwork [276]2 years ago

7 0

Answer:

What is freezing point?

A liquid's freezing point is determined at which it turns into a solid. Corresponding to the melting point, the freezing point often rises with increasing pressure. In the case of combinations and for some organic substances, such as lipids, the freezing point is lower than the melting point. The first solid which develops when a combination freezes often differs in composition from the liquid, and the development of the solid alters the composition of the remaining liquid, typically lowering the freezing point gradually. Utilizing successive melting and freezing to gradually separate the components, this approach is used to purify mixtures.

What is melting point?

The temperature at which a purified substance's solid and liquid phases may coexist in equilibrium is referred to as the melting point. A solid's temperature goes up when heat is added to it until the melting point is achieved. The solid will then turn into a liquid with further heating without changing temperature. Additional heat will raise the temperature of the liquid once all of the solid has melted. It is possible to recognize pure compounds and elements by their distinctive melting temperature, which is a characteristic number.

The difference between freezing point and melting point:

While a substance's melting point develops when it transforms from a solid to a liquid, a substance's freezing point happens when a liquid transforms into a solid when the heat from the substance is removed.
When the temperature rises, the melting point can be seen, and when the temperature falls, the freezing point can be seen.
When a solid reaches its melting point, its volume increases; meanwhile, when a liquid reaches its freezing point, its volume decreases.
While a substance's freezing point is not thought of as a distinctive attribute, its melting point is.
While external pressure is a significant component in freezing point, atmospheric pressure is a significant element in melting point.
Heat must be supplied from an outside source in order to reach the melting point for such a state shift. When a material is at its freezing point, heat is needed to remove it from the substance in order to alter its condition.

Reference: Berry, R. Stephen. "When the melting and freezing points are not the same." Scientific American 263.2 (1990): 68-75.

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viva [34]

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$v_{f2}$ =6.5% $v_{i1}$

Explanation:

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Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

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Differences between freezing point and melting point (Atleast 5 differences)​

Differences between freezing point and melting point (Atleast 5 differences)