First part concerning the signs.
The sign depends upon so many things; e.g., if we use Tf-Ti for delta T or just delta T and if we use heat lost = heat gained OR if we use heat lost + heat gained = 0. Frankly, I can get so confused with all of that that I prefer to use
mass water x specific heat water x (Tf-Ti) + Cp calorimeter x (Tf-Ti) = 0. THE SIGNS THEN TAKE CARE OF THEMSELVES.
I would point out two or three things.
1. The problem quotes 39.8o C but you used 39.6 for T in your calculations. That makes a small difference.
I'll do the others as a separate response.
Can you check what I did below? On the first problem, I am not sure if I am suppose to add a negative to the first part of the heat loss of water like so -[C(sp)m(w)deltaT(w)]. On the second problem, part(a), I am not sure if I am using the mass from the first problem, 135g, or obtain the mass by multipling 100 mL with the density of water at 45.3*C or 31.8*C, to solve for the heat gained by the solution. Also, I am not sure if I am using the heat of capacity of the calorimeter in the first question to solve part (b). Thanks for the help!
Consider a calorimeter into which 135 g of water at 45.3*C is added. The initial temperature of the calorimeter was 23.8*C and the equilibrium temperature of the water and the calorimeter is 39.8*C. Assuming there is no heat loss to the surroundings, what is the heat capacity of the calorimeter in J/*C?
Heat loss of H20=Heat gain of calorimeter
-[C(sp)m(w)deltaT(w)]=[C(p)deltaT(c)]
-[(4.184 J/g*C)(135g)(39.6*C-45.3*C)]=C(p)(39.6*C-23.8*C)
C(p)=2.0x10^2 J/*C
In another experiment, 11.6 g of a salt was dissolved in 100.0 mL of water contained in the above calorimeter, all at an initial temperature of 24.5*C. After dissolution was complete, the temperature was 31.8*C. The specific heat of the solution is 3.63 J/g*C and the molar mass of the salt is 153.5 g/mol. Assume there is no heat loss to the surroundings.
a)Calculate the heat gained by the solution
Q(solution)=CM(solution)deltaT(solution)
=(3.63 J/g*C)(11.6g +135g)(31.8*C-24.5*C)
=3.88x10^3 J
b)Calculate the heat gained by the calorimeter
Q(c)=C(p)deltaT(c)
=(2.0x10^2 J/*C)(39.6*C-23.8*C)
=3.2x10^3 J
c)Calculate the total heat gained by the system
(3.9x10^3 J) + (3.2x10^3 J) = 7.1 x 10^3 J
d)Calculate the heat of solution (deltaH) of the salt in J/mol
(7.1x10^3 J)/[(11.6)(1 mol/153.5g)]
=(7.1x10^3 J)/ .0756 mol = 9.4 x 10^4 J/mol
6 answers
Part a.
mass x specific heat x (Tf-Ti).
The 135 g H2O was in the other problem, not this one so forget it. The number you want here is, technically, 100.0 mL H2O x density water at 31.8 BUT since density is not given in the problem, my best guess is that the authors intended you to use 100.0 mL = 100.0 grams. So Q = 100.0 x 3.63 x (31.8-24.5) = ??
b.
Q calorimeter = Cp x (Tf-Ti). Again, you are using numbers from the first problem. Using from this problem Tf = 31.8 and Ti = 24.5.
c.
The total heat gained by the system must be the heat absorbed by the solution + the heat absorbed by the calorimeter or answer from a + answer from b.
d.
heat solution in j/mol = total heat/mols salt.
mols salt = 11.6/153.5
mass x specific heat x (Tf-Ti).
The 135 g H2O was in the other problem, not this one so forget it. The number you want here is, technically, 100.0 mL H2O x density water at 31.8 BUT since density is not given in the problem, my best guess is that the authors intended you to use 100.0 mL = 100.0 grams. So Q = 100.0 x 3.63 x (31.8-24.5) = ??
b.
Q calorimeter = Cp x (Tf-Ti). Again, you are using numbers from the first problem. Using from this problem Tf = 31.8 and Ti = 24.5.
c.
The total heat gained by the system must be the heat absorbed by the solution + the heat absorbed by the calorimeter or answer from a + answer from b.
d.
heat solution in j/mol = total heat/mols salt.
mols salt = 11.6/153.5
For part d, the heat of solution is a negative number since placing the salt in water made the T rise from 24.5 to 31.8, that makes it an exothermic reaction, and that means delta H is negative. It gives off heat.
Opps, there is a typo when I typed out the problem. The equilibrium temperature in the first problem is supposed to be 39.6*C.
For the 100 mL, I assume my teacher wants me to look up the denisty in the appendix of our lab notebook. Am I suppose to look up the density of water at 24.5*C, which is 0.997914g/ml?
Wouldn't the heat capacity of the calorimeter be different for part c) if I am not using 135 g water?
The one thing that troubles me the most is that the second problem refers to the water contained in the above calorimeter.
Thanks for the help DrBob222!
For the 100 mL, I assume my teacher wants me to look up the denisty in the appendix of our lab notebook. Am I suppose to look up the density of water at 24.5*C, which is 0.997914g/ml?
Wouldn't the heat capacity of the calorimeter be different for part c) if I am not using 135 g water?
The one thing that troubles me the most is that the second problem refers to the water contained in the above calorimeter.
Thanks for the help DrBob222!
My questions are in the previous reply. It would be great if you clear up those uncertanities. Thanks! So, with the corrections, I have this:
1) Consider a calorimeter into which 135 g of water at 45.3*C is added. The initial temperature of the calorimeter was 23.8*C and the equilibrium temperature of the water and the calorimeter is 39.6*C. Assuming there is no heat loss to the surroundings, what is the heat capacity of the calorimeter in J/*C?
Heat loss of H20=Heat gain of calorimeter
-[C(sp)m(w)deltaT(w)]=[C(p)deltaT(c)]
-[(4.184 J/g*C)(135g)(39.6*C-45.3*C)]=C(p)(39.6*C-23.8*C)
C(p)=2.0x10^2 J/*C
2) In another experiment, 11.6 g of a salt was dissolved in 100.0 mL of water contained in the above calorimeter, all at an initial temperature of 24.5*C. After dissolution was complete, the temperature was 31.8*C. The specific heat of the solution is 3.63 J/g*C and the molar mass of the salt is 153.5 g/mol. Assume there is no heat loss to the surroundings.
a)Calculate the heat gained by the solution
Q(solution)=CM(solution)deltaT(solution)
(3.63 J/g*C)(11.6g + (100ml)(0.997914g/ml))(31.8*C-24.5*C)=3.0x10^3 J
b)Calculate the heat gained by the calorimeter
Q(c)=C(p)deltaT(c)
=(2.0x10^2 J/*C)(31.8*C-24.5*C)
=4.5x10^3 J
c)Calculate the total heat gained by the system
(3.0x10^3 J) + (1.5x10^3 J) = 4.5 x 10^3 J
d)Calculate the heat of solution (deltaH) of the salt in J/mol
(4.5x10^3 J)/[(11.6)(1 mol/153.5g)]
=(4.5x10^3 J)/ .0756 mol = -6.0 x 10^4 J/mol
1) Consider a calorimeter into which 135 g of water at 45.3*C is added. The initial temperature of the calorimeter was 23.8*C and the equilibrium temperature of the water and the calorimeter is 39.6*C. Assuming there is no heat loss to the surroundings, what is the heat capacity of the calorimeter in J/*C?
Heat loss of H20=Heat gain of calorimeter
-[C(sp)m(w)deltaT(w)]=[C(p)deltaT(c)]
-[(4.184 J/g*C)(135g)(39.6*C-45.3*C)]=C(p)(39.6*C-23.8*C)
C(p)=2.0x10^2 J/*C
2) In another experiment, 11.6 g of a salt was dissolved in 100.0 mL of water contained in the above calorimeter, all at an initial temperature of 24.5*C. After dissolution was complete, the temperature was 31.8*C. The specific heat of the solution is 3.63 J/g*C and the molar mass of the salt is 153.5 g/mol. Assume there is no heat loss to the surroundings.
a)Calculate the heat gained by the solution
Q(solution)=CM(solution)deltaT(solution)
(3.63 J/g*C)(11.6g + (100ml)(0.997914g/ml))(31.8*C-24.5*C)=3.0x10^3 J
b)Calculate the heat gained by the calorimeter
Q(c)=C(p)deltaT(c)
=(2.0x10^2 J/*C)(31.8*C-24.5*C)
=4.5x10^3 J
c)Calculate the total heat gained by the system
(3.0x10^3 J) + (1.5x10^3 J) = 4.5 x 10^3 J
d)Calculate the heat of solution (deltaH) of the salt in J/mol
(4.5x10^3 J)/[(11.6)(1 mol/153.5g)]
=(4.5x10^3 J)/ .0756 mol = -6.0 x 10^4 J/mol
Opps, there is a typo when I typed out the problem. The equilibrium temperature in the first problem is supposed to be 39.6*C. That takes care of that.
For the 100 mL, I assume my teacher wants me to look up the density in the appendix of our lab notebook. Am I supposed to look up the density of water at 24.5*C, which is 0.997914g/ml?
If you are to go the density route, I think you want to look up the density at the FINAL temperature. The final T is the one at which equilibrium is reached so that will determine the final mass of water. Most problems of this type I have seen say to assume the solution has a density of 1.0; however, with 11.6 g of a salt and only 100.0 mL water, I can see that the density probably will NOT be 1.00 if that is all the water used. But see my next comment; I think you have closer to 235 g water. Even then, however, the water might be significantly different from a density of 1.00, expecially for accurate calculations.
Wouldn't the heat capacity of the calorimeter be different for part c) if I am not using 135 g water?
Heat capacity is heat capacity. The way I have interpreted the problem is that the 100.0 mL plus the salt is added to the existing 135 g water + the container of the calorimeter. Therefore, the heat capacity of the 135 g water + the container is taken into account by Cp*delta T (for the experiment, not delta T for the heat capacity determination) and you don't add in the extra water for the 100.0 mL + salt.
The one thing that troubles me the most is that the second problem refers to the water contained in the above calorimeter.
That's why I interpreted as I explained above. USUALLY, the heat capacity of a calorimeter is determined in this type experiment with a certain amount of water and that amount of water is used each time.
For the 100 mL, I assume my teacher wants me to look up the density in the appendix of our lab notebook. Am I supposed to look up the density of water at 24.5*C, which is 0.997914g/ml?
If you are to go the density route, I think you want to look up the density at the FINAL temperature. The final T is the one at which equilibrium is reached so that will determine the final mass of water. Most problems of this type I have seen say to assume the solution has a density of 1.0; however, with 11.6 g of a salt and only 100.0 mL water, I can see that the density probably will NOT be 1.00 if that is all the water used. But see my next comment; I think you have closer to 235 g water. Even then, however, the water might be significantly different from a density of 1.00, expecially for accurate calculations.
Wouldn't the heat capacity of the calorimeter be different for part c) if I am not using 135 g water?
Heat capacity is heat capacity. The way I have interpreted the problem is that the 100.0 mL plus the salt is added to the existing 135 g water + the container of the calorimeter. Therefore, the heat capacity of the 135 g water + the container is taken into account by Cp*delta T (for the experiment, not delta T for the heat capacity determination) and you don't add in the extra water for the 100.0 mL + salt.
The one thing that troubles me the most is that the second problem refers to the water contained in the above calorimeter.
That's why I interpreted as I explained above. USUALLY, the heat capacity of a calorimeter is determined in this type experiment with a certain amount of water and that amount of water is used each time.
0 answers