Answer:
A. 56 kJ
B. 75.8 kJ
C. 11 ˚C
Explanation:
A. The heat of vaporization, ∆Hvap = 40.7 kJ/mol, gives the amount of energy per mole of water required to vaporize water to steam. The molar mass of water is 18.02 g/mol.
Q = M·∆Hvap = (25 g)(mol/18.02g)(40.7 kJ/mol) = 56 kJ
B. Five steps are necessary in this process. First, the ice will be warmed to 0 °C, then melted to water. The water will be heated to 100 °C, then vaporized. Finally, the vapor will be heated from 100 °C to 105 °C.
We calculate the heat required to warm the ice from -4.0 °C to 0 °C:
Q₁ = mcΔt = (25 g)(2.06 J∙g⁻¹˚C⁻¹)(0 °C - (-4.0 °C)) = 206 J
Then we calculate the heat required to melt the ice to water:
Q₂ = M∙∆Hfus = (25 g)(mol/18.02 g)(6.02 kJ/mol) = 8.35 kJ
Then, we calculate the heat required to warm the water from 0 °C to 100 °C.
Q₃ = mcΔt = (25 g)(4.184 J∙g⁻¹˚C⁻)(100 °C - 0 °C) = 10460 J
Then we calculate the heat required to vaporize the water:
Q₄ = M∙∆Hvap = (25 g)(mol/18.02 g)(40.7 kJ/mol) = 56.5 kJ
Finally, the vapor is heated from 100 °C to 105 °C.
Q₅ = mcΔt = (25 g)(2.03 J∙g⁻¹˚C⁻)(105 °C - 100 °C) = 254 J
The total heat required is the sum of Q₁ through Q₅
Qtotal = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qtotal = (206 J)(1 kJ/1000J) + 8.35 kJ + (10460 J)(1 kJ/1000J) + 56.5 kJ + (254 J)(1 kJ/1000J)
Qtotal = 75.8 kJ
C. The heat required to melt the ice is provided by the water as it decreases in temperature.
First, we calculate the energy required to melt ice to water
Q = M∙∆Hfus = (8.32 g)(mol/18.02 g)(6.02 kJ/mol) = 2.779 kJ
There are at least two ways to solve this problem. Here, we will calculate the heat lost when all the water is brought to a temperature of 0 °C:
Q = mc∆t = (55 g)(4.184 J∙g⁻¹˚C⁻¹)(25 °C - 0°C) = 5753 J
We see that the water has enough energy to melt all of the ice. The residual heat energy of the water after melting all the ice is:
5753 J - (2.779 kJ)(1000J/kJ) = 2974 J
Now the problem becomes that we have (8.32 g + 55 g) = 63.32 g of water at 0 °C that will be raised to some final temperature by the residual heat of 2974 J:
Q = mcΔt ⇒ Δt = Q/(mc)
Δt = (2974 J) / (63.32 g)(4.184 J∙g⁻¹˚C⁻¹) = 11 ˚C
T(final) - T(inital) = 11 ˚C
T(final) = 11 ˚C + T(inital) = 11 ˚C + 0 ˚C = 11 ˚C
Thus, the final temperature will be 11 ˚C.
