How much energy is required to change a 40 g ice cube from ice at -25°C to steam at 105°C?
40 g = .04 kg
.04 [25 (sh ice) + heat of fusion + 100 (sh water) + heat of vaporization + 5 (sh water vapor at 1 atm) ]
I don't know if this is right that i pulled in..but i got it wrong. Please help.
4(2090)+(3.33⋅10^5 )+100(4186)+(2.26⋅10^6)+5(4186)
To calculate the energy required to change a substance from one state to another, you need to consider the following changes:
1. Heating the ice from -25°C (-248 K) to its melting point: The specific heat capacity of ice is 2.09 J/g°C. So, to heat the ice from -25°C to 0°C, you need to calculate the energy using the formula:
q = m × c × ΔT
where:
- q is the energy in joules (J),
- m is the mass of the ice cube in grams (40 g),
- c is the specific heat capacity of ice (2.09 J/g°C),
- ΔT is the change in temperature (0°C - (-25°C) = 25°C.
By plugging in these values:
q1 = 40 g × 2.09 J/g°C × 25°C = 2090 J.
2. Melting the ice at 0°C: To melt the ice, you need to provide additional energy known as the heat of fusion. The heat of fusion of ice is 333.55 J/g. The formula to calculate the energy required to melt the ice is:
q = m × ΔHf
where:
- q is the energy in joules (J),
- m is the mass of the ice cube in grams (40 g),
- ΔHf is the heat of fusion (333.55 J/g).
Plugging in these values:
q2 = 40 g × 333.55 J/g = 13342 J.
3. Heating the water from 0°C to 100°C: The specific heat capacity of water is 4.18 J/g°C. So, to heat the water from 0°C to 100°C, you need to calculate the energy using the formula:
q = m × c × ΔT
where:
- q is the energy in joules (J),
- m is the mass of the water in grams (40 g),
- c is the specific heat capacity of water (4.18 J/g°C),
- ΔT is the change in temperature (100°C - 0°C) = 100°C.
Plugging in these values:
q3 = 40 g × 4.18 J/g°C × 100°C = 16720 J.
4. Evaporating the water at 100°C to steam at 100°C: To convert water at its boiling point to steam, you need to provide additional energy known as the heat of vaporization. The heat of vaporization of water is 2260 J/g. The formula to calculate the energy required to vaporize the water is:
q = m × ΔHv
where:
- q is the energy in joules (J),
- m is the mass of the water in grams (40 g),
- ΔHv is the heat of vaporization (2260 J/g).
Plugging in these values:
q4 = 40 g × 2260 J/g = 90400 J.
Now, to get the total energy required, you simply add up all the individual energy values:
Total energy = q1 + q2 + q3 + q4 = 2090 J + 13342 J + 16720 J + 90400 J = 122552 J.
Therefore, the total energy required to change a 40 g ice cube from ice at -25°C to steam at 105°C is 122,552 J.