What total energy must be absorbed (from the environment) to change 1.00kg of ice at -30.0°C on a lake surface in February to water at 20.0°C in August?
Add the following heats..
heat ice from -30 to OC
melt ice at 0C
heat water from 0C to 20C
To calculate the total energy absorbed, we need to consider the three processes mentioned:
1. Heating the ice from -30°C to 0°C :
For this, we need to use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.
For ice, the specific heat is 2108 J/kg°C.
Q1 = (1.00 kg) * (2108 J/kg°C) * (0°C - (-30°C))
Q1 = 1.00 kg * 2108 J/kg°C * 30°C
Q1 = 63,240 J
2. Melting the ice at 0°C:
For this process, we need to use the formula Q = mL, where L is the heat of fusion for ice, which is 333.5 kJ/kg.
Q2 = (1.00 kg) * (333.5 kJ/kg)
Q2 = 333.5 kJ = 333,500 J
3. Heating the water from 0°C to 20°C:
For this process, we'll again use the formula Q = mcΔT. The specific heat of water is 4186 J/kg°C.
Q3 = (1.00 kg) * (4186 J/kg°C) * (20°C - 0°C)
Q3 = 1.00 kg * 4186 J/kg°C * 20°C
Q3 = 83,720 J
Now, let's sum up the energies for all three processes:
Total energy = Q1 + Q2 + Q3
Total energy = 63,240 J + 333,500 J + 83,720 J
Total energy = 480,460 J
So, the total energy absorbed is 480,460 Joules.
To calculate the total energy required to change 1.00 kg of ice at -30.0°C to water at 20.0°C, we need to sum up the energy required for each step.
1. Heating the ice from -30.0°C to 0°C:
The formula for calculating the energy required to heat a substance is Q = m × c × ΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For ice, the specific heat capacity varies slightly with temperature, but we can use an average value of 2.09 J/g°C.
To convert the mass to grams, we multiply 1.00 kg by 1000, which gives us 1000.00 g.
The change in temperature is 0°C - (-30.0°C) = 30.0°C.
Using the formula, we find Q1 = (1000.00 g) × (2.09 J/g°C) × (30.0°C).
2. Melting the ice at 0°C:
The heat required to melt ice is calculated using the formula Q = m × ΔHf, where ΔHf is the heat of fusion. For water, it is 334 J/g.
Using the same mass of 1000.00 g, we find Q2 = (1000.00 g) × (334 J/g).
3. Heating the water from 0°C to 20.0°C:
Using the same formula as step 1, with the specific heat capacity of liquid water being 4.18 J/g°C, we calculate Q3 = (1000.00 g) × (4.18 J/g°C) × (20.0°C).
Finally, we sum up all the calculated values to get the total energy required:
Total Energy = Q1 + Q2 + Q3.
Now you can substitute the values into the equations and perform the calculations to find the total energy.
To find the total energy that must be absorbed, we need to calculate the heat absorbed at each step and then add them together.
Step 1: Heating the ice from -30.0°C to 0°C
The specific heat capacity of ice is 2.09 J/g·°C, and we have 1.00 kg of ice.
Therefore, the heat absorbed in this step can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat absorbed, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = (1.00 kg) * (2.09 J/g·°C) * (0 - (-30.0°C))
Step 2: Melting the ice
The heat of fusion for ice is 334 J/g. Since we have 1.00 kg of ice, we multiply the heat of fusion by the mass to find the total heat absorbed in this step.
Q2 = (1.00 kg) * (334 J/g)
Step 3: Heating the water from 0°C to 20.0°C
The specific heat capacity of water is 4.18 J/g·°C, and we have 1.00 kg of water.
Using the same formula as before, we can calculate the heat absorbed in this step.
Q3 = (1.00 kg) * (4.18 J/g·°C) * (20.0°C - 0°C)
To find the total energy absorbed, we just need to add the heats together.
Total Heat Absorbed = Q1 + Q2 + Q3
Note: Make sure to convert the units appropriately to maintain consistency throughout the calculations.