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 59.4 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.
Solution to this problem
To calculate the total quantity of heat necessary to melt the ice and heat it to steam, we will break down the process into three parts:
1. Heating the ice from -50.0°C to 0°C
2. Melting the ice at 0°C
3. Heating the water from 0°C to 200°C and transforming it into steam
1. Heating the ice from -50.0°C to 0°C:
The specific heat capacity of ice is given as 2.06 J/g·◦C. We need to calculate the heat required to increase the temperature of the ice from -50.0°C to 0°C.
q1 = m * c1 * ΔT1
q1 = 59.4 g * 2.06 J/g·◦C * (0°C - (-50.0°C))
q1 = 59.4 g * 2.06 J/g·◦C * 50.0°C
q1 = 6093.6 J
So, the heat required to raise the temperature of the ice from -50.0°C to 0°C is 6093.6 J.
2. Melting the ice at 0°C:
To calculate the heat required to melt the ice, we use the molar heat of fusion for water, given as 6.02 kJ/mol. Since we know the mass of the ice, we need to convert it into moles.
moles of ice = mass of ice / molar mass of water
moles of ice = 59.4 g / (18.015 g/mol)
moles of ice ≈ 3.296 mol
q2 = moles of ice * molar heat of fusion
q2 = 3.296 mol * 6.02 kJ/mol * 1000 J/kJ
q2 = 19792.192 J
So, the heat required to melt the ice at 0°C is 19792.192 J.
3. Heating the water from 0°C to 200°C and transforming it into steam:
The specific heat capacity of water is given as 4.18 J/g·◦C. We need to calculate the heat required to heat the water from 0°C to 100°C and then to 200°C.
q3 = m * c2 * ΔT2
q3 = 59.4 g * 4.18 J/g·◦C * (100°C - 0°C)
q3 = 59.4 g * 4.18 J/g·◦C * 100°C
q3 = 248902.4 J
So, the heat required to heat the water from 0°C to 100°C is 248902.4 J.
Next, we use the molar heat of vaporization for water, given as 40.6 kJ/mol, to calculate the heat required to transform the water into steam.
q4 = moles of water * molar heat of vaporization
q4 = 3.296 mol * 40.6 kJ/mol * 1000 J/kJ
q4 = 133657.6 J
So, the heat required to transform the water into steam at 100°C is 133657.6 J.
Finally, we calculate the heat required to heat the steam from 100°C to 200°C:
q5 = m * c2 * ΔT3
q5 = 59.4 g * 2.03 J/g·◦C * (200°C - 100°C)
q5 = 59.4 g * 2.03 J/g·◦C * 100°C
q5 = 120081.6 J
So, the heat required to heat the steam from 100°C to 200°C is 120081.6 J.
Now, we can find the total quantity of heat necessary by summing up all the calculated values:
Total heat = q1 + q2 + q3 + q4 + q5
Total heat = 6093.6 J + 19792.192 J + 248902.4 J + 133657.6 J + 120081.6 J
Total heat ≈ 529527 J
Therefore, the total quantity of heat necessary to melt 59.4 g of ice at -50.0°C and heat it to steam at 200°C is approximately 529,527 J.
Watch the units. You omitted the specific heat of liquid water. It is 4.184 J/g*C. Also you should convert 59.4 g H2O to mols H2O. mols = g/molar mass = ? and use the appropriate value for mass H2O (either mols or grams) with the specific heat (either J/g or kJ/mol).
q1 = heat to move T of ice from -50 to 0 C.
q2 = heat to melt ice
q3 = heat to move T liquid water from 0 C to 100 C
q4 = heat to boil & vaporize all of the water
q5 = heat to move T of H2O vapor from 100 C to 200 C.
q total = q1 + q2 + q3 + q4 + q5
q1 =mass ice(grams) x specific heat ice (J/g) x (Tfinal-Tnitial) = q in J.
q2 = mass ice (mol) x sp.h. (kJ/mol) = q in kJ. Convert to J.
q3 = mass water (g) x sp.h. (J/g) x (Tfinal-Tinitial) = q in J.
q4 = mass water(mols) x heat vaporization (kJ/mol) = q in kJ. Convert to J.
q5 = mass water (g) x sp.h. steam (J/g) x (Tfinal-Tinitial) = q in J.
Post your complete work if you get stuck.
oops. q2 should be
q2 = mass ice (mol) x heat fusion (kJ/mol) = q in kJ. Convert to J.