What is the minimum amount of heat required to completely melt 20 g of ice at its melting point

The amount of solute required to form a saturated solution in a particular amount of solvent is the Concentration.

Heat of fusion is the amount of heat energy required to change the state of matter of a substance from a solid to a liquid. It's also known as enthalpy of fusion. Its units are usually Joules per gram (J/g) or calories per gram (cal/g). This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice.

  • Heat of fusion is the amount of energy in the form of heat needed to change the state of matter from a solid to a liquid (melting.)
  • The formula to calculate heat of fusion is: q = m·ΔHf
  • Note that the temperature does not actually change when matter changes state, so it's not in the equation or needed for the calculation.
  • Except for melting helium, heat of fusion is always a positive value.

What is the heat in Joules required to melt 25 grams of ice? What is the heat in calories?

Useful information: Heat of fusion of water = 334 J/g = 80 cal/g

In the problem, the heat of fusion is given. This isn't a number you're expected to know off the top of your head. There are chemistry tables that state common heat of fusion values.

To solve this problem, you'll need the formula that relates heat energy to mass and heat of fusion:
q = m·ΔHfwhere

q = heat energy

m = mass

ΔHf = heat of fusion

Temperature is not anywhere in the equation because it doesn't change when matter changes state. The equation is straightforward, so the key is to make sure you're using the right units for the answer.

To get heat in Joules:q = (25 g)x(334 J/g)q = 8350 JIt's just as easy to express the heat in terms of calories:

q = m·ΔHf

q = (25 g)x(80 cal/g)q = 2000 cal

Answer: The amount of heat required to melt 25 grams of ice is 8,350 Joules or 2,000 calories.

Note: Heat of fusion should be a positive value. (The exception is helium.) If you get a negative number, check your math.


The following solution is suggested to handle the subject “What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve“. Let’s keep an eye on the content below!

Question “What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve”

What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve

Answer

To answer this question, it is necessary to understand the water’s value.

Enthalpy for Fusion
,


DeltaH_f

.

You know that the enthalpy for fusion of a substance tells you how hot it is
Required To melt


“1 g”

Of


Ice At


0^@”C”

To


liquid At


0^@”C”

.

Simply put, the enthalpy for fusion of a substance tells you how hot it needs to fuse.

“1 g”

Water to be used for a
Solid

->

liquid
Phase change
.

The enthalpy for water fusion is roughly equal


DeltaH_f = 334″J”/”g”


http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

This means that you need to melt

“1 g”

Ice at

0^@”C”

Liquid water

0^@”C”

You must provide it with

“334 J”

Heat.

You have

“20.0 g”

You will need to have ice melting capability.

20

Heat up to ten times more More than you’d need


“1 g”

Ice.

You can therefore say that


20.0 color(red)(cancel(color(black)(“g ice”))) * “334 J”/(1color(red)(cancel(color(black)(“g ice”)))) = “6680 J”

Are required to melt

“20.0 g”

Ice at

0^@”C”

Liquid water

0^@”C”

.

Therefore,
(C)

“6680 J”

This is the answer.

Conclusion

Above is the solution for “What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve“. We hope that you find a good answer and gain the knowledge about this topic of chemistry.


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What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve