How much heat (energy) is required to convert 18.0 g (1.00 mole) of ice at -5.00oC to steam
at a temperature 111oC? (6)
all of these phases have different thermal coefficients
... heat ice to 0ºC
... melt ice
,,, heat water to 100ºC
... boil water
... heat steam to 111ºC
Answer
To calculate the amount of heat required to convert a substance from one state to another, we need to consider the following steps:
1. Heating the ice from -5.00°C to 0.00°C: The heat required for this step can be calculated using the formula Q = m × C × ΔT, where Q is the heat energy in joules, m is the mass in grams, C is the specific heat capacity of the substance, and ΔT is the change in temperature. For ice, the specific heat capacity is 2.09 J/g°C. So, the heat required to raise the temperature of 18.0 g of ice from -5.00°C to 0.00°C can be calculated as Q1 = 18.0 g × 2.09 J/g°C × (0.00 – (-5.00))°C.
2. Melting the ice at 0.00°C: The heat required for this step can be calculated using the formula Q = m × ΔHf, where Q is the heat energy in joules, m is the mass in grams, and ΔHf is the heat of fusion (also known as the latent heat of fusion) for the substance. For ice, the heat of fusion is 333.5 J/g. So, the heat required to melt 18.0 g of ice at 0.00°C can be calculated as Q2 = 18.0 g × 333.5 J/g.
3. Heating the water from 0.00°C to 100.00°C: The heat required for this step can be calculated using the formula Q = m × C × ΔT, where Q is the heat energy in joules, m is the mass in grams, C is the specific heat capacity of the substance, and ΔT is the change in temperature. For water, the specific heat capacity is 4.18 J/g°C. So, the heat required to raise the temperature of the 18.0 g of water from 0.00°C to 100.00°C can be calculated as Q3 = 18.0 g × 4.18 J/g°C × (100.00 – 0.00)°C.
4. Vaporizing the water at 100.00°C: The heat required for this step can be calculated using the formula Q = m × ΔHv, where Q is the heat energy in joules, m is the mass in grams, and ΔHv is the heat of vaporization (also known as the latent heat of vaporization) for the substance. For water, the heat of vaporization is 2260 J/g. So, the heat required to vaporize the 18.0 g of water at 100.00°C can be calculated as Q4 = 18.0 g × 2260 J/g.
5. Heating the steam from 100.00°C to 111.00°C: The heat required for this step can be calculated using the formula Q = m × C × ΔT, where Q is the heat energy in joules, m is the mass in grams, C is the specific heat capacity of the substance, and ΔT is the change in temperature. For steam, the specific heat capacity is 2.03 J/g°C. So, the heat required to raise the temperature of the 18.0 g of steam from 100.00°C to 111.00°C can be calculated as Q5 = 18.0 g × 2.03 J/g°C × (111.00 – 100.00)°C.
Now, we can calculate the total heat required to convert 18.0 g (1.00 mole) of ice at -5.00°C to steam at a temperature 111.00°C by summing up the individual heats calculated in each step:
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
Calculate each of the individual heats as described above and sum them up to find the total heat required.