Heat is added to 0.77 kg of ice at -25 °C. How many kilocalories are required to change the ice to steam at 131 °C?
I don't even know where to begin with this
add the heat to heat the ice to 0C
Then the heat to melt the ice at 0C
then to heat the water from 0C t0 100C
then to convert the water to steam at 100C
then to heat the steam from 100C to 131C
add them all up.
To answer this question, you will need to calculate the amount of heat required to change the ice to water at 0 °C, then the heat required to change the water to steam at 100 °C, and finally the heat required to raise the temperature of the steam from 100 °C to 131 °C.
Let's break it down step by step:
Step 1: Heat required to change ice to water
- The heat required to change the state of a substance is given by the formula: Q = m * L, where Q is the heat, m is the mass, and L is the latent heat of fusion.
- For ice, the latent heat of fusion (L) is 334 kilojoules per kilogram (kJ/kg).
- First, convert the mass from kg to grams: 0.77 kg = 770 grams.
- The heat required to change the ice to water is: Q1 = m * L = 770 g * 334 kJ/kg = 257,380 kJ.
Step 2: Heat required to change water to steam
- The heat required to change water to steam is given by the formula: Q = m * L, where Q is the heat, m is the mass, and L is the latent heat of vaporization.
- For water, the latent heat of vaporization (L) is 2256 kilojoules per kilogram (kJ/kg).
- The mass of the water remains the same at 770 grams.
- The heat required to change the water to steam is: Q2 = m * L = 770 g * 2256 kJ/kg = 1,735,520 kJ.
Step 3: Heat required to raise the temperature of the steam
- The heat required to raise the temperature of a substance is given by the formula: Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- For water or steam, the specific heat capacity (c) is approximately 4.18 kilojoules per kilogram per degree Celsius (kJ/(kg·°C)).
- The mass of the steam remains the same at 770 grams.
- ΔT is the change in temperature, which is from 100 °C to 131 °C, so ΔT = 131 °C - 100 °C = 31 °C.
- The heat required to raise the temperature of the steam is: Q3 = m * c * ΔT = 770 g * 4.18 kJ/(kg·°C) * 31 °C = 98,446.6 kJ.
Finally, add up all three heats to find the total heat required:
Total heat = Q1 + Q2 + Q3 = 257,380 kJ + 1,735,520 kJ + 98,446.6 kJ.
Note that this answer is given in kilojoules, but if you want the answer in kilocalories, you can convert it by dividing by 4.18 (since 1 kilocalorie is equal to 4.18 kilojoules).