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Ch.5 - Thermochemistry

Chapter 5, Problem 60c

Under constant-volume conditions, the heat of combustion of naphthalene (C10H8) is 40.18 kJ/g. A 2.50-g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to 28.83 °C. (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

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Welcome back everyone. The heat program of a sample of benzene was measured at constant volume using a calorie meter. The first sample of 3.70 g of benzene was burned with an observed temperature increase of 8.93 degrees Celsius. The second sample was 9.332 g of benzene. If the water was lost during changing the sample, which of the following will be observed. So what we want to first think of is our relation of our heat exchange in the reaction expressed as our entropy or delta H. Where specifically we are focusing on the heat exchange of our component being our samples. And this is going to be related to the heat of our water and the heat of our calorie meter. Recall that heat is expressed by the term Q. And so we can further define this to emphasize that the heat of our samples or the component is equal to the mass of the water times the molar specific heat capacity of water represented by C. Sub P. Multiplied by the difference in temperature delta T. For water and then added to the specific heat of our calorie meter, times the delta T. Or difference in temperature that the calorie meter experiences. And we want to recognize that our units for entropy are expressed in joules per mole, which we would further emphasize as jules times the molar mass of our substance divided by grams of our sample. Now going back to what the prompt tells us, we're told that for sample one we have a mass of 3.70 g of benzene C six H six. And for sample too, We have a higher mass of 9.32 g of Benzene. Now we want to make note of the fact that according to the prompt, water is lost during changing the sample. And so with this in mind, when we get to sample too, we know that water is lost and so therefore the heat per gram of our sample is going to ultimately decrease. Since we see that there's a direct relationship of the heat of our component to the heat of water. And so what this means is that we can also say we have a decrease in jewels program of our sample whereas on the converse for sample one, water is not lost as yet. So we have a higher value of heat per gram for sample one. And so with this outlined, we can confirm that the only correct choice to complete this example is going to be Choice B which states that the heat per gram of the first sample will be greater than the heat program of the second sample. Note that the specifically capacity is not going to change with regards to our samples because specifically capacity is a is a constant. So it would only change if we changed our substance. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video
Related Practice
Textbook Question

A 2.20-g sample of phenol (C6H5OH) was burned in a bomb calorimeter whose total heat capacity is 11.90 kJ/°C. The temperature of the calorimeter plus contents increased from 21.50 to 27.50 °C. (b) What is the heat of combustion per mole of phenol?

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Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C6H5O6) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (a) What is the total heat capacity of the calorimeter?

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Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C6H5O6) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

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Textbook Question

Consider the following hypothetical reactions: A → B ΔHI = +60 kJ B → C ΔHII = -90 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

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Textbook Question

Calculate the enthalpy change for the reaction P4O6(s) + 2 O2(g) → P4O10(s) given the following enthalpies of reaction: P4(s) + 3 O2(g) → P4O6(s) ΔH = -1640.1 kJ P4(s) + 5 O2(g) → P4O10(s) ΔH = -2940.1 kJ

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Textbook Question

From the enthalpies of reaction 2 C(s) + O2(g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O2(g) + 4 H2(g) → 2 CH3OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H2(g) → CH3OH(g)

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