Question

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)?

Answer 1

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.