Answer the following using the following information: ∆Hfus=6.02 kJ/mol; ∆Hvap= 40.7 kJ/mol; specific heat of water is 4.184 J/g∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of water vapor is 2.03 J/g∙˚C.

A. How much heat is required to vaporize 25 g of water at 100˚C?

B. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)?

C. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?

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.