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Chapter 5: Problem 53

The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) is \(2.22 \mathrm{~J} / \mathrm{g}\) -K. \((\mathbf{a})\) How many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from 10.0 to \(25.0^{\circ} \mathrm{C} ?(\mathbf{b})\) 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 g of octane from 10.0 to 25.0掳C is 2664 J. \(b)\) Increasing the temperature of 1 mole of octane by the same amount requires more heat than increasing the temperature of 1 mole of water.

Step by step solution

01

Part (a): Calculate the heat required to raise the temperature of octane

We are given the mass of octane (\(m = 80.0\) g), specific heat of octane (\(c = 2.22\) J/g-K), and the initial (\(T_1 = 10.0掳C\)) and final (\(T_2 = 25.0掳C\)) temperatures. We can now calculate the heat required using the formula: \(q = mc螖T\). 1. Calculate the change in temperature: \(螖T = T_2 - T_1 = 25.0掳C - 10.0掳C = 15.0掳C\). 2. Calculate the heat required: \(q = (80.0\text{ g})(2.22\text{ J/g-K})(15.0\text{ K}) = 2664\text{ J}\). So, the heat required to raise the temperature of 80.0 g of octane from 10.0 to 25.0掳C is 2664 J.
02

Part (b): Compare the heat required for 1 mole of octane and 1 mole of water

First, we need to find the molar mass of octane \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\): 1. Molar mass of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\): \((8 \times 12.01\text{ g/mol}) + (18 \times 1.01\text{ g/mol}) = 114.22\text{ g/mol}\) Now, let's assume the change in temperature for both octane and water is \(螖T\): 2. Heat required for 1 mole of octane (\(q_1\)): \(q_1 = (1\text{ mol})(114.22\text{ g/mol})(2.22\text{ J/g-K})螖T = 253.47螖T\text{ J}\) 3. Heat required for 1 mole of water (\(q_2\)): \(q_2 = (1\text{ mol})(18.02\text{ g/mol})(4.18\text{ J/g-K})螖T = 75.33螖T\text{ J}\) Comparing the heat required for 1 mole of octane and 1 mole of water: 4. Check if \(q_1 > q_2\): \((253.47螖T\text{ J}) > (75.33螖T\text{ J})\) Since the heat required for 1 mole of octane (\(q_1\)) is greater than the heat required for 1 mole of water (\(q_2\)), increasing the temperature of 1 mole of octane by the same amount requires more heat than increasing the temperature of 1 mole of water.

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Key Concepts

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

Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. It can occur in various ways, such as conduction, convection, or radiation. Understanding how heat transfer works is crucial in calculating the energy required for changing the temperature of substances.

In the context of our exercise, we're calculating how much heat is needed to raise the temperature of octane. This involves using the specific heat capacity formula:
  • \[ q = mc螖T \]

Here,
  • \( q \) represents the heat energy (in joules),
  • \( m \) is the mass (in grams),
  • \( c \) is the specific heat capacity (in J/g-K),
  • \( 螖T \) is the change in temperature (in Kelvin or Celsius).
Heat transfer calculations help us understand how much energy we need to apply or remove to achieve a desired temperature change. This is essential for various practical applications, whether we are cooking food or engineering sophisticated thermal management systems.
Molar Mass
Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a given substance. It is usually expressed in units of grams per mole (g/mol). Calculating molar mass is crucial when dealing with reactions and energy changes for a given amount of substance.

In our problem, we calculate the molar mass of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), by adding up the atomic masses of all atoms in its molecular formula:
  • \[ (8 \times 12.01\, \text{g/mol}) + (18 \times 1.01\, \text{g/mol}) = 114.22\, \text{g/mol} \]
This molar mass calculation helps us in converting moles to grams, thus facilitating the comparison of energy needs for different substances. When comparing the energy required to heat one mole of octane versus water, we first establish the molar mass to estimate how much each mole weighs in grams. This provides a foundational step in determining the heat needed for each mole separately.
Temperature Change
Temperature change (\(螖T\)) is the difference between the final and initial temperatures of a substance. It is a critical variable in calculating heat transfer, as it directly influences how much heat energy is absorbed or released.

In our example, the temperature change is calculated as:
  • \[ 螖T = T_2 - T_1 = 25.0掳C - 10.0掳C = 15.0掳C \]
This change is crucial because it tells us how much the temperature has shifted, setting the stage for calculating the heat needed using the equation \( q = mc螖T \).

Whether you're warming water or cooling down machinery, understanding the concept of temperature change helps you quantify the energy shifts occurring within a given system. The larger the temperature change, the more energy will be involved in the heat transfer process, assuming mass and specific heat are constant.

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Most popular questions from this chapter

Assume that 2 moles of water are formed according to the following reaction at constant pressure \((101.3 \mathrm{kPa})\) and constant temnerature \((298 \mathrm{~K});\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the pressure-volume work for this reaction. (b) Calculate \(\Delta E\) for the reaction using your answer to (a).

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{I}_{2}(s)\) or \(1 \mathrm{~mol} \mathrm{I}_{2}(g)\) at the same temperature, (b) \(2 \mathrm{~mol}\) of iodine atoms or \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2},(\mathbf{c}) 1 \mathrm{~mol} \mathrm{I}_{2}(g)\) and \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(2 \mathrm{~mol} \mathrm{HI}(g)\) at \(25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(300^{\circ} \mathrm{C}\).

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of \(\mathrm{AgCl} . (\mathbf{c})\) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

The complete combustion of methane, \(\mathrm{CH}_{4}(g)\), to form \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(890 \mathrm{~kJ}\) of heat per mole of \(\mathrm{CH}_{4}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and \((\mathbf{b}) 10.4 \mathrm{~g}\) of \(\mathrm{KCl} .(\mathbf{c})\) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.
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