Winter ice: heat of fusion

As the ice melts away in the sun, the tutor entertains the heat involved.

The heat of fusion is the amount of energy, per gram, that is required to melt solid to liquid without increasing its temperature. For ice, it’s 334J/g.

The tutor wonders how quickly the ice on the pavement melts in the sunlight on a day like today.

We’ll start with some assumptions. Let’s imagine we have a layer of ice 0.1cm (aka 1mm or 0.001m) thick over the pavement. Then the volume of ice per square metre is 1mx1mx0.001m = 0.001m³. The density of ice is 917kg/m³, so the mass of ice per square metre of pavement would be 0.917kg = 917g. Furthermore, we’ll assume the temperature is hovering around 0°C.

To find the amount of heat required to melt 917g of ice, we multiply it by ice’s heat of fusion, earlier said to be 334J/g. 917gx334J/g=306278J.

The wattage of sunlight reaching the pavement, on a day like today, might be 341W/m².

The time for the ice to melt under the assumptions above is t=306278÷341=898s: 15 minutes.

All the best this holiday season:)

Source:

hypertextbook.com

itacanet.org

Hebden, James A. Chemistry: Theory and Problems, book two. Toronto: McGraw-Hill,
   1980.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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