Question

How many grams of ice at 0 degrees Celsius and 101.3 kPa could be melted by the addition of 0.400 KJ of heat?

Answer 1

This question can be solved by knowing the enthalpy of fusion of water, so divide the given amount of heat with the enthalpy of fusion of water, this will give the number of moles of water that will be melted. Now, this number of moles of water can be converted into grams by multiplying it by the molecular mass of water.

Complete answer:
We are given that 0.400 KJ of heat is applied to the ice for the melting of ice. The conditions that are given are 0 degrees Celsius and 101.3 kPa pressure, these are standard conditions, so there will change in the effect of melting due to these conditions.
We can calculate the number of moles of water melted by dividing the given amount of heat with the energy of fusion of water. The enthalpy of fusion of water is 6007 KJ/ mol. The number of moles will be:
$\dfrac{0.400}{6007}=0.066$
So, the number of moles of water melted is 0.066 moles.
Now, this number of moles of water can be converted into grams by multiplying it by the molecular mass of water. The molecular mass is 18 g /mol. The grams of water will be:
0.066 x 18 = 1.198 grams or 1.2 grams
So, 1.2 grams of water will be melted when 0.400 KJ of heat is applied.

Note:
This way of solution only can be done if the conditions are standard, i.e., 0 degree Celsius and 101.3 kPa pressure. If the conditions are different then the value of water melted will also change.