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The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), is \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). (a) How many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from \(10.0\) to \(25.0^{\circ} \mathrm{C}\) ? (b) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) by the same amount?

Short Answer

Expert verified
a) The heat required to raise the temperature of \(80.0 \mathrm{~g}\) of octane from \(10.0\) to \(25.0^{\circ} \mathrm{C}\) is \(2664 \mathrm{~J}\). b) Increasing the temperature of \(1\mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) (octane) will require more heat than increasing the temperature of \(1\mathrm{~mol}\) of \(\mathrm{H}_{2}\mathrm{O}(l)\) (water) by the same amount.

Step by step solution

01

Calculate the change in temperature

We have given that the initial temperature is \(10.0^{\circ} \mathrm{C}\) and the final temperature is \(25.0^{\circ} \mathrm{C}\). The change in temperature is calculated as: \(\Delta T = T_{f} - T_{i}\) \(\Delta T = 25.0 - 10.0 = 15.0 ^{\circ} \mathrm{C}\)
02

Calculate the heat required for octane

Now we can use the formula \(q = mC\Delta T\) to find the heat required to raise the temperature of \(80.0 \mathrm{~g}\) of octane. The specific heat for octane is \(2.22 \mathrm{~J}/\mathrm{g} \cdot \mathrm{K}\), the mass is \(80.0 \mathrm{~g}\), and the change in temperature is \(15.0 ^{\circ} \mathrm{C}\). Plug these values into the formula: \(q = (80.0 \mathrm{~g})(2.22 \mathrm{~J}/\mathrm{g} \cdot \mathrm{K})(15.0 \mathrm{~K})\) \(q = 2664 \mathrm{~J}\) a) The heat required is \(2664\mathrm{~J}\).
03

Compare the heat requirements for 1 mol of octane and 1 mol of water

To compare which substance requires more heat, we need to calculate the heat capacity (specific heat multiplied by the molar mass) for both substances. The specific heat of water is \(4.18 \mathrm{~J}/\mathrm{g}\cdot \mathrm{K}\). The molar mass of octane is \(12.01 \cdot 8+1.01 \cdot 18 = 114.23 \mathrm{~g/mol}\) and the molar mass of water is \(1.01 \cdot 2 +16.00 = 18.02 \mathrm{~g/mol}\). Heat capacity of octane: \((2.22 \mathrm{~J}/\mathrm{g} \cdot \mathrm{K})(114.23 \mathrm{~g/mol}) = 253.6 \mathrm{~J}/\mathrm{mol}\cdot \mathrm{K}\) Heat capacity of water: \((4.18 \mathrm{~J}/\mathrm{g}\cdot \mathrm{K})(18.02 \mathrm{~g/mol}) = 75.3 \mathrm{~J}/\mathrm{mol}\cdot \mathrm{K}\) b) Increasing the temperature of \(1\mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) (octane) will require more heat than increasing the temperature of \(1\mathrm{~mol}\) of \(\mathrm{H}_{2}\mathrm{O}(l)\) (water) by the same amount.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Required for Temperature Change
When we talk about heating a substance, what we're really speaking about is the energy required for a temperature change. This concept is essential in understanding how substances absorb heat. The amount of energy required to raise the temperature of a specific amount of a substance is calculated using the formula:

\[ q = mC\Delta T \]

Where:\
    \
  • \(q\) represents the heat energy in joules (J),\
  • \(m\) represents the mass of the substance in grams (g),\
  • \(C\) is the specific heat capacity (J/g·K), and\
  • \(\Delta T\) is the change in temperature in Kelvin (K) or degrees Celsius (°C).\
\
It's worth noting that the temperature change (\(\Delta T\)) is the same whether you measure it in Kelvin or degrees Celsius as both scales have the same size 'degree'. The specific heat capacity, \(C\), is a substance's characteristic property; it tells us how much heat is needed to raise the temperature of one gram of the substance by one degree Celsius (or one Kelvin).

In the case of our octane example, we applied this formula and found that it takes 2664 J to raise the temperature of 80 grams of octane by 15°C. This tells us that different substances require different amounts of energy to achieve the same temperature change, which is a direct consequence of their varying specific heats.
Comparing Heat Capacities
Now let's delve into the concept of comparing heat capacities of different substances. When we say 'heat capacity', we refer to the amount of heat needed to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). Heat capacity is a bulk property, and it's related to specific heat but it takes into account the amount of substance (in moles).

To calculate it, we multiply the specific heat (C) by the molar mass (M) of the substance:

\[ Heat \, Capacity = C \times M \]

Thus, for our octane, with a specific heat of 2.22 J/g·K and a molar mass of 114.23 g/mol, its heat capacity is lower compared to water, which means that octane would require more energy to warm up the same number of moles by the same temperature change. This difference in heat capacities signifies how substances absorb and retain heat differently and is pivotal for applications ranging from engineering to cooking.
Enthalpy and Chemistry Thermodynamics
Lastly, let's discuss the broader context of enthalpy and chemistry thermodynamics. In chemistry, enthalpy is the total heat content of a system and it's a key concept in thermodynamics. When we calculate the energy required to trigger a temperature change, we’re often considering changes in enthalpy, denoted by \(\Delta H\). Enthalpy change reflects the heat absorbed or released by a system at constant pressure.

For instance, in a reaction where chemical bonds are formed or broken, the change in enthalpy can tell us if the reaction is exothermic (releases heat) or endothermic (absorbs heat). This ties back to our earlier discussions because knowing a substance’s specific heat can help in calculating these enthalpy changes during chemical reactions.

Understanding how to manipulate and measure heat and temperature changes is not just academic; it's crucial for various applications in everyday life, industry, and environmental science. By mastering these principles, students can better grasp the often invisible, yet immensely powerful, forces that drive chemical transformations.

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Most popular questions from this chapter

5.104 We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) $$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$ \begin{array}{ll} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-890.3 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) & \Delta H^{\circ}=-136.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-3120.8 \mathrm{~kJ} \end{array} $$

From the enthalpies of reaction $$ \begin{gathered} 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) \quad \Delta H=-221.0 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{CH}_{3} \mathrm{OH}(g) \quad \Delta H=-402.4 \mathrm{~kJ} \\ \text { calculate } \Delta H \text { for the reaction } \\ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) \end{gathered} $$

At the end of 2012 , global population was about \(7.0\) billion people. What mass of glucose in kg would be needed to provide \(1500 \mathrm{cal} /\) person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-2803 \mathrm{~kJ} \end{aligned} $$

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanestburning octanes is a compound called 2,3,4-trimethylpentane, which has the following structural formula: CC(C)C(C)C(C)C The complete combustion of one mole of this compound to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) leads to \(\Delta H^{\circ}=-5064.9 \mathrm{~kJ} / \mathrm{mol}\). (a) Write a balanced equation for the combustion of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\). (b) By using the information in this problem and data in Table 5.3, calculate \(\Delta H_{f}^{\circ}\) for \(2,3,4\)-trimethylpentane.

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