Answer: The ice absorb 6671.1 kJ of thermal energy.
Explanation:
The conversions involved in this process are :
[tex]0.00^0C=273K[/tex]
[tex]:H_2O(s)(273K)\rightarrow H_2O(l)(273K)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=n\times \Delta H_{fusion}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of ice = 20.0 kg = [tex]20.0\times 10^3g[/tex] (1kg=1000g)
n = number of moles of ice= [tex]\frac{\text{Mass of ice}}{\text{Molar mass of water}}=\frac{20.0\times 10^3g}{18g/mole}=1.11\times 10^3mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=1.11\times 10^3mole\times 6010J/mole[/tex]
[tex]\Delta H=6671100J=6671.1kJ[/tex] (1 kJ = 1000 J)
Therefore, the enthalpy change is, 6671.1 kJ
Answer:
Explanation:
Heat Absorbed = mL
where,
L = latent heat of fusion = [tex]334kg/kJ[/tex]
Therefore,
Heat absorbed = [tex]20 * 334[/tex]
Heat absorbed = [tex]6680KJ[/tex]
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