Calculate the amount of heat required to melt 2108 g of water. The enthalpy of fusion of water is ΔHfus=6.010 kJ/mol.

Answers

Answer 1

since the unit for the heat of fusion is kJ/mol, you're going to have to convert the grams into moles in order to cancel out the unit. After that, you can solve like normal.

Calculate The Amount Of Heat Required To Melt 2108 G Of Water. The Enthalpy Of Fusion Of Water Is Hfus=6.010
Answer 2

Specific heat capacity of a substance is the amount of heat required to raise the temperature by one degree Celsius of one gram of a substance. Therefore, the amount of heat required to melt 2108 g of water is 703.7kJ.

What is Enthalpy?

Enthalpy term is basically used in thermodynamics to show the overall energy that a matter have. Mathematically, Enthalpy is directly proportional to specific heat capacity of a substances.

Mathematically,

q = n ×ΔH

where

q = amount of heat=?

ΔH = enthalpy=6.010 kJ/mol

n = no of moles

n = w / M.M

w = given mass

M.M = molar mass

moles of water=2108 g ÷18g/mol

                             =117.1 mol

substituting all the given values in the above equation, we get

q =6.010 kJ/mol ×117.1 mol

    =703.7kJ

Therefore, the amount of heat required to melt 2108 g of water is 703.7kJ.

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