how much energy is required to change 40 g of ice at -10 C to steam at 110 C?
This is a three part question.
First part is to calculate the amount of energy necessary to turn -10 C ice into 0 C water. Next is calculating the energy to convert 0 C water into 100 C steam. Then calculating the amount of energy to raise the steam 10 C to 110 C.
Use
Q=mcΔT
Q is heat energy
m is the mass of the thing
c is the specific heat capacity constant which is different depending on the material
ΔT is the change in temperature caused by the heat energy (or vice versa).
You also need to account for the Latent Heat of Fusion for Ice and Latent Heat of Vaporization for Steam, which is why phase changes are important.
In addition, the specific heat capacity value is different for each stage of water.
Part 1:
Q = 40g*(2.108 j/g*K)*(10)
Amount of heat energy necessary to go from -10 C ice to 0 C ice. Now add the latent head of fusion times the amount of mass to the Q value you just calculated. (334 j/g*40 g).
Do the same for the other parts of the questions.
Best solution thanks
Well, if you ask me, turning that ice into steam sounds like quite an ambitious project! So, let's do some calculations and see how much energy we'll need to pull it off, shall we?
To change ice at -10°C into water at 0°C, we'll need to use the specific heat capacity of ice, which is about 2.09 J/g°C. Given that we have 40 grams of ice and we want to raise its temperature by 10°C, we'll need:
40 g × 2.09 J/g°C × 10°C = 836 J
Now, once we've transformed the ice into water at 0°C, we'll need to use the specific heat capacity of water, which is approximately 4.18 J/g°C, to heat it up to 100°C. So, for that, we calculate:
40 g × 4.18 J/g°C × 100°C = 16,720 J
After reaching 100°C, we need to take into account the heat of vaporization to convert water into steam. For water, the heat of vaporization is approximately 2260 J/g. Thus, to turn all the water into steam, we'll need:
40 g × 2260 J/g = 90,400 J
Last but not least, we come to bringing the steam from 100°C to 110°C. Using the specific heat capacity of steam, which is around 2.03 J/g°C, we have:
40 g × 2.03 J/g°C × 10°C = 812 J
Now, let's add up all those values:
836 J + 16,720 J + 90,400 J + 812 J = 108,768 J
So, it seems we'll need a whopping 108,768 Joules of energy to change 40 grams of ice at -10°C to steam at 110°C. It's no joke, but hey, at least we'll get to see some amazing transformations!
To calculate the energy required to change 40 g of ice at -10 °C to steam at 110 °C, we need to consider the different stages of the phase change. The energy required can be calculated using the formula:
Q = m * (h_f + ΔT * C)
Where:
Q = energy in Joules
m = mass in grams
h_f = heat of fusion (energy required to change ice to water) in Joules/gram
ΔT = change in temperature in degrees Celsius
C = specific heat capacity in Joules/gram-degree Celsius
Let's break it down step-by-step:
1. Calculate the energy required to change the ice at -10 °C to water at 0 °C:
Q1 = m * (h_f + ΔT * C)
Since the ice is at -10 °C and we need to heat it up to 0 °C:
ΔT1 = (0 °C - (-10 °C)) = 10 °C
Using the specific heat capacity of ice (C = 2.09 J/g-°C) and the heat of fusion (h_f = 333.5 J/g), we can calculate Q1:
Q1 = 40 g * (333.5 J/g + 10 °C * 2.09 J/g-°C)
2. Calculate the energy required to change the water at 0 °C to steam at 100 °C:
Q2 = m * (h_v + ΔT * C)
Since we start at 0 °C and need to reach 100 °C:
ΔT2 = 100 °C - 0 °C = 100 °C
Using the specific heat capacity of water (C = 4.18 J/g-°C) and the heat of vaporization (h_v = 2260 J/g), we can calculate Q2:
Q2 = 40 g * (2260 J/g + 100 °C * 4.18 J/g-°C)
3. Calculate the energy required to change the water (steam) at 100 °C to steam at 110 °C:
Q3 = m * ΔT * C
Since we start at 100 °C and need to reach 110 °C:
ΔT3 = 110 °C - 100 °C = 10 °C
Using the specific heat capacity of steam (C = 2.03 J/g-°C), we can calculate Q3:
Q3 = 40 g * 10 °C * 2.03 J/g-°C
4. Calculate the total energy:
Total Energy = Q1 + Q2 + Q3
By substituting the respective values into the formulas, you can calculate the total energy required step-by-step.
To determine the energy required to change the state of a substance from solid to liquid, and from liquid to gas, we can use the concept of specific heat and latent heat.
First, let's calculate the energy required to change 40 g of ice at -10°C to water at 0°C (the melting point of ice):
1. Calculate the energy required to change the ice from its initial temperature (-10°C) to 0°C using the specific heat capacity of ice. The specific heat capacity of ice is 2.09 J/g°C.
Energy = mass × specific heat capacity × temperature change
Energy = 40 g × 2.09 J/g°C × (0°C - (-10°C))
2. Next, calculate the energy required to melt the ice at 0°C.
The latent heat of fusion for water is 334 J/g.
Energy = mass × latent heat of fusion
Energy = 40 g × 334 J/g
Now, let's calculate the energy required to change the water from 0°C to steam at 100°C:
3. Calculate the energy required to raise the temperature of water from 0°C to 100°C using the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C.
Energy = mass × specific heat capacity × temperature change
Energy = 40 g × 4.18 J/g°C × (100°C - 0°C)
4. Finally, calculate the energy required to vaporize the water at 100°C.
The latent heat of vaporization for water is 2260 J/g.
Energy = mass × latent heat of vaporization
Energy = 40 g × 2260 J/g
Total energy required = Sum of the energies calculated in steps 1, 2, 3, and 4.
Note: As the question asks for the energy required to reach 110°C, instead of 100°C, we need to account for the additional energy required to heat the steam from 100°C to 110°C using the specific heat capacity of steam, which is 2.03 J/g°C. You can calculate this additional energy by multiplying the mass of steam by the specific heat capacity of steam and the temperature change.
Remember to convert the temperatures to Kelvin (K) for the temperature differences (e.g., -10°C + 273.15 = 263.15 K) in the equations above.
By following these steps and performing the calculations, you can find the exact amount of energy required to change 40 g of ice at -10°C to steam at 110°C.