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Molar heat of fusion of ice?

Updated: 8/10/2023
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9y ago

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First you need to know at what temp. is the ice?

The latent heat of fusion for ice at 32 degrees F is 144 but per pound.

This will change 1 pound of ice at 32 degrees F into water at 32 degrees F.
It is defined as the amount of heat required for the 1 mole of ice to bring a change in its state, that is, from solid state to liquid state. It is also known as enthalpy of fusion, specific melting point or latent heat of fusion of ice. The particular temperature at which there is a change in the state of the ice is known as the melting point of ice.

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6y ago
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11y ago

It is defined as the amount of heat required for the 1 mole of ice to bring a change in its state, that is, from solid state to liquid state. It is also known as enthalpy of fusion, specific melting point or latent heat of fusion of ice. The particular temperature at which there is a change in the state of the ice is known as the melting point of ice.

Quantitatively the heat of fusion for water is:

79.72 cal/g

or

334.0 kJ/kg

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14y ago

Purpose: To Calculate the molar enthalpy for the fusion of ice. Materials: * 100-200 ml sytrofoam coffee cup and lid * 100 ml graduated cylinder. * thermometer * electronic scale or triple beam balance * ice * warm water (50°C) * paper towel Procedure: * Mass a coffee cup and lid. * Heat water in a 250 ml beaker to approximately 50°C. * Fill a sytrofoam coffee cup with 100 ml of heated water measured with the graduated cylinder * Take the temperature of the water. * Mass the coffee cup, water and lid. * Obtain some crushed ice. (Approximately 15 - 20 grams (two handfuls). * Dry the ice with paper towel to remove the sweat on their surface. * Record the temperature of the water in the coffee cup (this is temperature initial) * Place the coffee cup, lid and water on the scale and add enough ice to increase the mass Approximately 15 grams.(Note the mass does not have to increase exactly 15 grams). Place the lid on the sytrofoam cup and record the final mass. * Stir the ice until it all melts and record the final temperature in the sytrofoam cup. Data: Mass of coffee cup, lid, water and ice. Mass of Coffee cup, lid and water Mass of Coffee cup and lid Mass of water Mass of Ice Temperature of water (temperature before) Temperature of water after ice melted (temperature after) Calculations: # Calculate the mass of water in the calorimeter in kilograms. # Calculate the mass of ice added to the calorimeter in kilograms. # Calculate the moles of ice that was added (mass of ice in grams ÷ molar mass of water) = n. # Calculate the temperature change in the water in the calorimeter (temperature before - temperature after). # Calculate the temperature change that happened to the ice after it melted (this is the change from 0 oC to the final temperature of the ice water mixture). # The heat that the water in the calorimeter lost was absorbed by the ice in first melting, and then later in the ice warming up to the final temperature of the water in the calorimeter. The formulas for calculating these changes are given below. * Heat lost by the calorimeter Q calorimeter = m * c *t ** Q = heat in KJ ** m = mass of water in calorimeter in Kilograms (#1 above) ** c = specific heat capacity of water (4.19 J/g * oC) ** t = temperature change in calorimeter.(#4 above) * Heat gained by the ice in melting is Q melting of ice = H * n ** Q = heat in KJ ** H = molar enthalpy of fusion of ice ** n = moles of ice (#3 above) * Heat gained by the ice as it warms up to the final temperature in the calorimeter is Q ice water = m* c *t ** Q = heat in KJ ** m = mass of melted ice (same as the mass of the ice) (#2 above) ** c = specific heat capacity of melted ice (same as water = 4.19 J/g * oC) ** t = temperature change that occurred in the melted ice (#5 above ) Combined the formula is as follows: Heat change in calorimeter=Heat needed to melt ice+Heat absorbed as ice warmed upm*c*t= H*n+m*c*t 7. Plug in values from above into this formula and calculate the molar heat of fusion of ice ( H )

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9y ago

The molar enthalpy fusion of ice is larger than the enthalpy of many other solids. The amount of energy needed to melt one mole of a substance is its molar enthalpy of fusion.

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The amount of heat necessary to change 1 kg of a solid into a liquid at the same temperature is called the?

molar heat of fusion


The molar heats of sublimation and fusion of iodine are 62.3kjmol and 15.3kjmol respectively calculate the molar heat of vaporization of liquid iodide?

It is a known fact : Molar heat of sublimation = molar heat of fusion + molar heat of vaporization so, molar heat of vaporization = molar heat of sublimation - molar heat of fusion Mv = 62.3 kJ/mol - 15.3 kJ/mol Mv = 47 kJ/mol.


What is the molar heat of fusion of water?

The specific latent heat of fusion of water is 334 kJ/kg. Ice melts at 0 degrees Celsius and boils at 100 degrees Celsius.


What is the molar heat of fusion of ethanol?

4.931 kj/mol


What is the relation of the molar heat of fusion to the molar heat of vaporization?

Molar heat of fusion: the heat (enthalpy, energy) needed to transform a solid in liquid (expressed in kJ/mol). Molar heat of vaporization: the heat (enthalpy, energy) needed to transform a liquid in gas (expressed in kJ/mol).


A sentence for heat of fusion?

Because of the heat of fusion the ice is now water


How much heat is requires to convert 0.3 kilogram of ice at 0 C to water at the same temperature?

The molar heat of fusion of water in J / g is 334. To find the heat required to convert 0.3 kg, use the equation: heat of fusion * mass = heat required. It would require 100.2 kJ.


How do you calculate the molar heat of fusion?

Use Einstein's Theory of Special Relativity


How are the heat of fusion of ice and heat of fusion of water similar?

due to the anomalous behaviour of water.....


The molar heat of fusion for water is 6.008 kJmol What quantity of heat energy is released when 253 g of liquid water freezes?

When the molar enthalpy of fusion for water is 6.008 kJ/mol, there is 84.4 kJ released when 253 grams of liquid water freezes. 84.4 kJ


What is the value for the latent heat of fusion of ice?

First you need to know at what temp. is the ice?The latent heat of fusion for ice at 32 degrees F is 144 but per pound.This will change 1 pound of ice at 32 degrees F into water at 32 degrees F.It is defined as the amount of heat required for the 1 mole of ice to bring a change in its state, that is, from solid state to liquid state. It is also known as enthalpy of fusion, specific melting point or latent heat of fusion of ice. The particular temperature at which there is a change in the state of the ice is known as the melting point of ice.


What is the scientific name for ice forming and melting?

For forming it is Heat of (Fusion) and for melting its Heat of (Vaporization).