Need help on these chemistry math problems.....

M

MBH_14

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1. 24.22 g of ice at -13.8°C is placed in 48.97 g of water at 97.7°C in a perfectly insulated vessel. Assume that the molar heat capacities for H2O(s) and H2O(l) are 37.5 J/K/mol and 75.3 J/K/mol, respectively, and the molar enthalpy of fusion for ice is 6.01 kJ/mol. Calculate the final temperature. (You must answer in Kelvin, not °C.)

Hint Given (even with the hint i am still confused): You can prove that there will be no ice present in the final state by comparing the heat needed to raise the temperature of the ice to 0 degrees and to melt it with the heat that would be lost by the initial amount of liquid water as its temperature is lowered to the freezing point. To determine Tf, equate the sum of the heat needed to raise the ice to 0 degrees plus the heat needed to melt it plus the heat needed to further raise the temperature of the melted ice/water to the unknown final temperature with the heat lost by the initially liquid water as its temperature drops to the final temperature. Solve for Tf.

2. A 70 kg person uses 193.0 kJ of energy to walk a kilometer. This energy comes from "burning" glucose, but only about 30.0% of the heat of combustion of glucose can be used for propulsion. The rest is used for other body functions or is wasted as heat. Assuming that a sugar-coated cereal contains 37.0% sugar (which can be considered glucose) and no other energy source, what mass of cereal will provide enough energy to walk 1.2 km.

Out of the 15 problems i had, these were the only 2 i messed up one. Can someone please show me how to do them? Thank you for your time and help!
 
1. 24.22 g of ice at -13.8°C is placed in 48.97 g of water at 97.7°C in a perfectly insulated vessel. Assume that the molar heat capacities for H2O(s) and H2O(l) are 37.5 J/K/mol and 75.3 J/K/mol, respectively, and the molar enthalpy of fusion for ice is 6.01 kJ/mol. Calculate the final temperature. (You must answer in Kelvin, not °C.)

Hint Given (even with the hint i am still confused): You can prove that there will be no ice present in the final state by comparing the heat needed to raise the temperature of the ice to 0 degrees and to melt it with the heat that would be lost by the initial amount of liquid water as its temperature is lowered to the freezing point. To determine Tf, equate the sum of the heat needed to raise the ice to 0 degrees plus the heat needed to melt it plus the heat needed to further raise the temperature of the melted ice/water to the unknown final temperature with the heat lost by the initially liquid water as its temperature drops to the final temperature. Solve for Tf.
Click to expand...


We’re adding energy to ice to melt it and raise it to some temperature, Tf. We’re getting that energy from hot water, so it cools to Tf. Simplistically:

Ice + q (heat energy) => Tf <= Hot water – q (heat energy).

Since we’re in an insulated vessel, the q added to the ice equals the q taken from the hot water. The hint means we can start by assuming all the ice will melt. We need two calculations to find the energy required to melt the ice. One calc is to find the energy rqd to bring the ice to 0 degrees C (or 273.15 degrees K; but you can use 273K; it is probably best to work the whole problem in Kelvin). It is still solid after this first amount of energy is added:

q1 = s1*m1*(t2 – t1)

(In this case, specific heat, s1 = 37.5 J/K/mol; m1 = number of moles of ice; t2 = 0 + 273;
t1 = -13.8 + 273 Kelvin. You need to first calculate the number of moles of ice in 24.22g.)

The second calc is to find the energy rqd to melt the ice. The temp is still 0 degrees C (273K), but a phase change occurs, going from solid to liquid:

q2 = k*m1, where k is the heat of fusion of water: k = 6.01 kJ/mol.

Next, a final amount of energy transfer is rqd to raise the 0 degree C liquid water up to Tf:

q3 = s2*m1*(t2 – t1)

(In this case, specific heat, s2 = 75.3 J/K/mol; t2 = Tf; t1 = 0 degrees C (or 273K)).

The total energy gained by the ice mass is

Q1 = q1 + q2 + q3

The energy calc for the hot water being cooled is

Q2 = s2*m2*(t2 – t1) = s2*m2*((97.7 + 273) – Tf)

Q1 = -Q2

Therefore, we (you) can combine the equations and solve.
 
thanks a lot for your help wjm!! Any takers for the second one? I could really use some help on how to go about solving that one. Thanks again for the help!
 
Have you considered trying a chemistry help forum?

Just an idea, but math teachers don't always know industry-specific formulae and such for non-math subjects, so for non-math questions, you might have better luck on a topic-appropriate site.

Eliz.
 
stapel,

funny you said that, i did try that, and hopefully they can help me out. Thanks for all your guys help!
 
For question 2....

193 KJ for 1 km

231.6 KJ for 1.2 Km

at a 30 % efficiency the energy burned becomes

231.6 / 0.30 = 772 KJ for the 1.2 Km

To do the mass part you will need to have some density data of the ceral and sugar to find the mass and also the energy release of sugar per mass(J/gram)
With the J/g , find the mass of sugar required to provide 772 KJ, then increase this mass(divide by 0.37) to make it a full ceral mass.
 
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