To calculate the amount of heat energy required to convert 100 g of ice at 0 degrees Celsius into water at 50 degrees Celsius, we need to consider two separate processes:
- Heating the ice from 0 degrees Celsius to its melting point (0 degrees Celsius).
- Heating the water from its melting point (0 degrees Celsius) to 50 degrees Celsius.
Let's break down the calculations:
- Heating the ice to its melting point: The heat energy required can be calculated using the specific heat capacity equation:
Q1 = m * c1 * ΔT1
Where: Q1 is the heat energy (in joules), m is the mass of the ice (100 g), c1 is the specific heat capacity of ice (2.09 J/g°C), ΔT1 is the change in temperature (melting point of ice - initial temperature).
The melting point of ice is 0 degrees Celsius, so ΔT1 = 0 - 0 = 0 degrees Celsius.
Q1 = 100 g * 2.09 J/g°C * 0°C = 0 joules
Therefore, no heat energy is required to heat the ice to its melting point.
- Heating the water from the melting point to 50 degrees Celsius: The heat energy required can be calculated using the specific heat capacity equation:
Q2 = m * c2 * ΔT2
Where: Q2 is the heat energy (in joules), m is the mass of the water (100 g), c2 is the specific heat capacity of water (4.18 J/g°C), ΔT2 is the change in temperature (final temperature - melting point of ice).
The final temperature is 50 degrees Celsius, so ΔT2 = 50 - 0 = 50 degrees Celsius.
Q2 = 100 g * 4.18 J/g°C * 50°C = 20900 joules
Therefore, the amount of heat energy required to convert 100 g of ice at 0 degrees Celsius into water at 50 degrees Celsius is 20900 joules.
Please note that this calculation does not include the heat energy required for the phase change from solid (ice) to liquid (water) at the melting point, as it is already accounted for in the specific heat capacity of water.