Question

Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0°C when it was delivered. As it melted, how much thermal energy did the ice absorb?

Answer 1

Answer: The ice absorb 6671.1 kJ of thermal energy.

Explanation:

The conversions involved in this process are :

$0.00^0C=273K$

$:H_2O(s)(273K)\rightarrow H_2O(l)(273K)$

Now we have to calculate the enthalpy change.

$\Delta H=n\times \Delta H_{fusion}$

where,

$\Delta H$ = enthalpy change = ?

m = mass of ice = 20.0 kg  = $20.0\times 10^3g$    (1kg=1000g)

n = number of moles of ice= $\frac{\text{Mass of ice}}{\text{Molar mass of water}}=\frac{20.0\times 10^3g}{18g/mole}=1.11\times 10^3mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=1.11\times 10^3mole\times 6010J/mole$

$\Delta H=6671100J=6671.1kJ$     (1 kJ = 1000 J)

Therefore, the enthalpy change is,  6671.1 kJ