How much heat is necessary to change 430 grams of ice -9 Celsius to water at 20 Celsius.
To calculate the amount of heat required to change ice at -9 degrees Celsius to water at 20 degrees Celsius, you can follow these steps:
1. Determine the heat required to raise the temperature of the ice from -9 degrees Celsius to 0 degrees Celsius.
- The specific heat capacity of ice is 2.09 J/g°C.
- The mass of the ice is given as 430 grams.
- The temperature change is 0 - (-9) = 9 degrees Celsius.
- So, the heat required for this step is: Q1 = (mass)*(specific heat capacity)*(temperature change).
2. Determine the heat required to change the ice at 0 degrees Celsius to water at 0 degrees Celsius (melting).
- The heat of fusion (or heat required for melting) for ice is 334 J/g.
- The mass remains the same at 430 grams.
- So, the heat required for this step is: Q2 = (mass)*(heat of fusion).
3. Determine the heat required to raise the temperature of the water from 0 degrees Celsius to 20 degrees Celsius.
- The specific heat capacity of water is 4.18 J/g°C.
- The mass remains the same at 430 grams.
- The temperature change is 20 - 0 = 20 degrees Celsius.
- So, the heat required for this step is: Q3 = (mass)*(specific heat capacity)*(temperature change).
4. Calculate the total heat required by summing up all three steps:
Total heat = Q1 + Q2 + Q3.
Now, let's calculate the values:
Q1 = (430 g) * (2.09 J/g°C) * (9°C) = 8603.7 J
Q2 = (430 g) * (334 J/g) = 144020 J
Q3 = (430 g) * (4.18 J/g°C) * (20°C) = 359240 J
Total heat = Q1 + Q2 + Q3
Total heat = 8603.7 J + 144020 J + 359240 J
Total heat = 511863.7 J
Therefore, the amount of heat necessary to change 430 grams of ice at -9 degrees Celsius to water at 20 degrees Celsius is approximately 511,863.7 Joules.
heat ice to 0ºC
melt ice
heat water to 20ºC