How much heat energy is needed to transform 125g of ice at 0 degrees Celsius to warm water at 80 degrees Celsius?
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To calculate the amount of heat energy needed to transform ice at 0 degrees Celsius into warm water at 80 degrees Celsius, you would need to consider two separate processes: the heat energy required to change the ice to water at 0 degrees Celsius, and the heat energy required to warm the water from 0 degrees Celsius to 80 degrees Celsius.
First, let's calculate the energy required to change the ice to water. This process is known as the heat of fusion, which is the amount of heat energy required to change a substance from a solid to a liquid without changing its temperature. For water, the heat of fusion is 334 J/g. So, to change 125 g of ice at 0 degrees Celsius to water at 0 degrees Celsius, we would need:
Energy = mass * heat of fusion
Energy = 125 g * 334 J/g
Energy = 41,750 J
Next, let's calculate the energy needed to warm the water from 0 degrees Celsius to 80 degrees Celsius. The heat required to raise the temperature of a substance is known as the specific heat capacity. For water, the specific heat capacity is approximately 4.18 J/g°C. Therefore, to warm 125 g of water from 0 degrees Celsius to 80 degrees Celsius, we would need:
Energy = mass * specific heat capacity * temperature change
Energy = 125 g * 4.18 J/g°C * (80°C - 0°C)
Energy = 52,250 J
Adding both energies together, the total amount of heat energy needed to transform 125g of ice at 0 degrees Celsius to warm water at 80 degrees Celsius is:
Total Energy = Energy to change ice to water + Energy to warm the water
Total Energy = 41,750 J + 52,250 J
Total Energy = 94,000 J
Therefore, 94,000 Joules (J) of heat energy is needed for this transformation.