To determine the amount of heat required to heat the ice, we need to consider the different steps involved:
1. Heating the ice from -5.0 degrees Celsius to 0 degrees Celsius to melt it.
2. Heating the resulting water from 0 degrees Celsius to its final temperature of 40.0 degrees Celsius.
Let's calculate the amount of heat required for each step:
1. Heating the ice to 0 degrees Celsius:
The specific heat of ice is given as 2.1 J/g°C, which means it takes 2.1 joules of heat to raise the temperature of 1 gram of ice by 1 degree Celsius.
To raise the temperature of 25.0 g of ice by 5.0 degrees Celsius, we use the formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = 25.0 g * 2.1 J/g°C * 5.0 °C
Q1 = 262.5 joules
Therefore, 262.5 joules of heat are needed to heat the ice from -5.0 degrees Celsius to 0 degrees Celsius.
2. Melting the ice at 0 degrees Celsius:
The heat of fusion for water is the amount of heat required to convert 1 gram of ice at 0 degrees Celsius to water at 0 degrees Celsius. It is given as 334 J/g.
To determine the amount of heat required to melt the ice, we use the formula:
Q2 = m * Hf
where Q2 is the heat energy, m is the mass, and Hf is the heat of fusion.
Q2 = 25.0 g * 334 J/g
Q2 = 8,350 joules
Therefore, 8,350 joules of heat are needed to melt the ice.
3. Heating the resulting water to 40.0 degrees Celsius:
Now that we have water at 0 degrees Celsius, we need to heat it to the final temperature of 40.0 degrees Celsius.
Using the specific heat of water, which is generally 4.18 J/g°C, we can calculate the amount of heat required:
Q3 = m * c * ΔT
Q3 = 25.0 g * 4.18 J/g°C * 40.0 °C
Q3 = 4,180 joules
Therefore, 4,180 joules of heat are needed to heat the water from 0 degrees Celsius to 40 degrees Celsius.
To find the total amount of heat required, we need to add together the heat required for each step:
Total heat required = Q1 + Q2 + Q3
Total heat required = 262.5 J + 8,350 J + 4,180 J
Total heat required = 12,792.5 joules
Therefore, approximately 12,792.5 joules of heat are needed to heat 25.0 g of ice at -5.0 degrees Celsius to a final temperature of 40.0 degrees Celsius, considering the state of water at 40.0 degrees Celsius.