Heat of Fusion Example Problem - Melting Ice

Heat of fusion is the amount of heat energy required to change the state of matter of a substance from a solid to a liquid. Its also known as enthalpy of fusion. Its units are usually Joules per gram (J/g) or calories per gram (cal/g). This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice. Key Takeaways: Heat of Fusion for Melting Ice Heat of fusion is the amount of energy in the form of heat needed to change the state of matter from a solid to a liquid (melting.)The formula to calculate heat of fusion is: q m ·ÃŽâ€HfNote that the temperature does not actually change when matter changes state, so its not in the equation or needed for the calculation.Except for melting helium, heat of fusion is always a positive value. Example Problem What is the heat in Joules required to melt 25 grams of ice? What is the heat in calories? Useful information: Heat of fusion of water 334 J/g 80 cal/g Solution ï » ¿In the problem, the heat of fusion is given. This isnt a number youre expected to know off the top of your head. There are chemistry tables that state common heat of fusion values. To solve this problem, youll need the formula that relates heat energy to mass and heat of fusion:q m ·ÃŽâ€Hfwhereq heat energym massΔHf heat of fusion Temperature is not anywhere in the equation because it doesnt change when matter changes state. The equation is straightforward, so the key is to make sure youre using the right units for the answer. To get heat in Joules:q (25 g)x(334 J/g)q 8350 JIts just as easy to express the heat in terms of calories:q m ·ÃŽâ€Hfq (25 g)x(80 cal/g)q 2000 calAnswer: The amount of heat required to melt 25 grams of ice is 8,350 Joules or 2,000 calories. Note: Heat of fusion should be a positive value. (The exception is helium.) If you get a negative number, check your math.