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.
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
1 answer
1 answer