How many grams of ice at -11.2 degrees C can be completely converted to liquid at 29.8 degrees C if the available heat for this process is 4.12×10^3 kJ?
For ice, use a specific heat of 2.01 J/(g*degrees C) and delta H_fus = 6.01 kJ/mol.
How much heat is needed to raise the temperature of ice at -11.2 C to 0 C? That will x grams x 2.01 J/g*C x 11.2 = 22.51x.
How much heat is needed to melt the ice?
That will be xgrams x 333.6 J/g = 333.6x.
How much heat is needed to raise the temperature from 0C to 29.8.
That will be x grams x 4.186 x (29.8) = 124.7x
Total = 124.7x + 22.51x + 333.6x = ?x
?x = 4120J
Solve for x.
To solve this problem, we need to calculate the energy required to heat the ice from -11.2 degrees Celsius to 0 degrees Celsius, then the energy required to melt the ice at 0 degrees Celsius, and finally the energy required to heat the resulting liquid water from 0 degrees Celsius to 29.8 degrees Celsius. Let's break down the calculations step by step:
1. Calculate the energy required to heat the ice from -11.2 degrees Celsius to 0 degrees Celsius:
- The specific heat capacity of ice is 2.01 J/(g*°C).
- The mass of ice is unknown but can be denoted as 'm' grams.
- The temperature change is (0 - (-11.2)) = 11.2 degrees Celsius.
The energy required can be calculated using the formula:
Energy = mass * specific heat capacity * temperature change
Plugging in the values:
Energy1 = m * 2.01 J/(g*°C) * 11.2 °C
2. Calculate the energy required to melt the ice at 0 degrees Celsius:
- The enthalpy of fusion (delta H_fus) for ice is 6.01 kJ/mol.
- We need to convert 'm' grams of ice to moles of ice.
- The molar mass of water (H2O) is 18.015 g/mol.
The moles of ice can be calculated using the formula:
Moles of ice = mass of ice / molar mass of ice
Plugging in the values:
Moles = m / 18.015 g/mol
The energy required can be calculated using the formula:
Energy2 = Moles * delta H_fus
3. Calculate the energy required to heat the resulting liquid water from 0 degrees Celsius to 29.8 degrees Celsius:
- The specific heat capacity of liquid water is 4.18 J/(g*°C).
- The mass of liquid water is the same as the mass of ice, denoted as 'm' grams.
- The temperature change is (29.8 - 0) = 29.8 degrees Celsius.
The energy required can be calculated using the formula:
Energy3 = m * 4.18 J/(g*°C) * 29.8 °C
Now, let's add up the three energies and set it equal to the available heat (4.12×10^3 kJ):
Energy1 + Energy2 + Energy3 = 4.12×10^3 kJ
Solve this equation to find the value of 'm', which represents the mass of ice in grams that can be completely converted to liquid water at 29.8 degrees Celsius.