To calculate the heat absorbed by 100g of ice as it changes from -20°C to water at 0°C, you need to consider the specific heat capacity of ice and water, as well as the heat of fusion.
The specific heat capacity of ice is approximately 2.09 J/g°C, which means it takes 2.09 joules of energy to raise the temperature of 1 gram of ice by 1 degree Celsius.
To raise the temperature of the ice from -20°C to 0°C, you need to add:
Q1 = mass × specific heat capacity × change in temperature = 100g × 2.09 J/g°C × (0°C - (-20°C)) = 100g × 2.09 J/g°C × 20°C = 4180 J
So, 4180 joules of energy are needed to raise the temperature of 100g of ice from -20°C to 0°C.
Next, we need to consider the heat of fusion, which is the amount of heat required to convert a substance from a solid to a liquid without changing its temperature. For water, the heat of fusion is approximately 334 J/g.
To convert the ice at 0°C to water at 0°C, you need to add:
Q2 = mass × heat of fusion = 100g × 334 J/g = 33400 J
Therefore, the total heat absorbed by 100g of ice as it changes from -20°C to water at 0°C is:
Total heat = Q1 + Q2 = 4180 J + 33400 J = 37580 J
So, 100g of ice absorbs approximately 37,580 joules of heat to become water at 0°C.