Consider the combustion of a 0.30 g sample of butter in a bomb calorimeter having a heat capacity of 2.67 kJ/°C. If the temperature of the calorimeter increases from 23.5°C to 27.3°C, what is the energy of combustion (in kJ/g) of butter.

06A

Question

Consider the combustion of a 0.30 g sample of butter in a bomb calorimeter having a heat capacity of 2.67 kJ/°C. If the temperature of the calorimeter increases from 23.5°C to 27.3°C, what is the energy of combustion (in kJ/g) of butter.

06A

BlackTom AI · Accepted Answer

To determine the energy of combustion per gram, start by calculating the total heat absorbed by the calorimeter using q = C × ΔT. Here, C = 2.67 kJ/°C and ΔT = 27.3°C − 23.5°C = 3.8°C, so q = 2.67 × 3.8 ≈ 10.146 kJ. This is the energy released by the combustion of the butter, but conventionally we assign a negative sign to combustion energy because it is energy that leaves the system; thus the calorimeter records q = −10.146 kJ for the 0.30 g sample. Now convert this to per-gram energy: energy per gram = (−10.146 kJ) / 0.30 g ≈ −33.82 kJ/g, which rounds to about −34 kJ/g.

Option by option:
- Option 1: “–34 kJ/g” — This matches the calculation above when rounding to the nearest whole number, consistent with the sign convention for exothermic combustion. The magnitude is in the expected range given the data.
- Option 2: “–2.3 kJ/g” — This would imply a much smaller heat release per gram. It would require either a far smaller ΔT, a much smaller calorimeter heat capacity, or a much larger sample mass than given; none of these align with the provided values.
- Option 3: “–0.21 kJ/g” — This is even smaller per-gram energy and indicates an error by a factor of about 100 or a misapplication of units. It suggests dividing by 150 g instead of 0.30 g, which is not the case here.
- Option 4: “–3.0 kJ/g” — This value is still an order of magnitude off from the correct per-gram energy. It could arise from miscalculating ΔT or misplacing a decimal, but the numbers given do not support this result.

In summary, the proper approach is to first find the total heat change from the calorimeter, then divide by the mass to obtain energy per gram, keeping track of the sign to reflect exothermic combustion. The calculated value around −33.8 kJ/g corresponds to the standard rounding shown in Option 1, while the other options reflect common mistakes such as incorrect ΔT, incorrect division, or ignoring the negative sign for exothermic processes.

Consider the combustion of a 0.30 g sample of butter in a bomb calorimeter having a heat capacity of 2.67 kJ/°C. If the temperature of the calorimeter increases from 23.5°C to 27.3°C, what is the energy of combustion (in kJ/g) of butter. 06A单项选择题

类似问题

Q3 V3If the temperature of 100 mL of solution increased by 3 Kelvin (K) during the reaction and the specific heat capacity of the solution is 4.18 J/g°C, how much heat energy was absorbed?

Q1 V2The reaction between 100 mL of 0.10 M HCl and 100 mL of 0.10 M NaOH causes a temperature rise of 1.60 oC.What would be the temperature rise for a reaction of 100 mL of 0.10 M HCl with 50 ml of 0.10 M NaOH?HCl(aq) + NaOH(aq) [math: →]\ce{ \bond{->} } NaCl(aq) + H2O(l)

A sample comprising 0.0100 mol ethane (C2H6) is combusted in excess oxygen inside a bomb calorimeter that has a calorimeter constant of 440 J K–1. The calorimeter temperature increases from 22.0 to 26.6 °C. What is the change in internal energy accompanying this reaction?

What three factors do you need to know in order to calculate the heat change of a chemical reaction in a calorimeter?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单