Given that the specific heat capacities of ice

and steam are 2.06 J/g ·
◦C and 2.03 J/g·◦C, respectively, calculate the total quantity of
heat necessary to melt 19.2 g of ice at −50.0◦C and heat it to steam at 200◦C. The molar heats of fusion and vaporization for water are
6.02 kJ/mol and 40.6 kJ/mol, respectively. Assume water freezes at 0◦C and boils at 100◦C.
Answer in units of J.

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To calculate the total quantity of heat necessary to melt the ice and heat it to steam, we need to break down the process into three steps:

Step 1: Heating the ice from -50.0°C to 0°C
Step 2: Melting the ice at 0°C
Step 3: Heating the water from 0°C to 200°C and then boiling it to steam at 100°C

Step 1: Heating the ice from -50.0°C to 0°C
To calculate the heat required for this step, we will use the formula:

q = m * c * ΔT

where:
q = heat energy (in J)
m = mass of the substance (in g)
c = specific heat capacity (in J/g·°C)
ΔT = change in temperature (in °C)

For the ice, the specific heat capacity is given as 2.06 J/g·°C. The mass of ice is 19.2 g and the change in temperature is 0°C - (-50.0°C) = 50.0°C.

q1 = 19.2 g * 2.06 J/g·°C * 50.0°C
q1 = 1975.04 J

Step 2: Melting the ice at 0°C
To calculate the heat required for this step, we will use the formula:

q = m * ΔHf

where:
q = heat energy (in J)
m = mass of the substance (in g)
ΔHf = molar heat of fusion (in J/g)

The molar heat of fusion for water is given as 6.02 kJ/mol. To convert it to J/g, we need to divide the molar heat by the molar mass of water, which is approximately 18.0 g/mol.

ΔHf = (6.02 kJ/mol) / (18.0 g/mol)
ΔHf = 0.3344 kJ/g = 334.4 J/g

q2 = 19.2 g * 334.4 J/g
q2 = 6436.48 J

Step 3: Heating the water from 0°C to 200°C and then boiling it to steam at 100°C
To calculate the heat required for this step, we will use the formula:

q = m * c * ΔT

For the water, the specific heat capacity is given as 4.18 J/g·°C. The mass of water is still 19.2 g and the change in temperature is 200°C - 0°C = 200°C.

q3 = 19.2 g * 4.18 J/g·°C * 200°C
q3 = 15835.2 J

Since the water boils at 100°C, we need to add the molar heat of vaporization to convert water from liquid to steam. The molar heat of vaporization for water is given as 40.6 kJ/mol.

ΔHv = 40.6 kJ/mol = 40,600 J/mol

Now, we need to determine the number of moles of water present. We can use the formula:

moles = mass / molar mass

moles = 19.2 g / 18.0 g/mol
moles = 1.0667 mol

q4 = 1.0667 mol * 40,600 J/mol
q4 = 43,316 J

Finally, to get the total quantity of heat necessary, we sum up all the heat values from the three steps:

Total heat = q1 + q2 + q3 + q4
Total heat = 1975.04 J + 6436.48 J + 15835.2 J + 43316 J
Total heat = 67,562.72 J

Therefore, the total quantity of heat necessary to melt 19.2 g of ice at -50.0°C and heat it to steam at 200°C is 67,562.72 J.