Problem 53
Question
The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(I),\) is \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K} .(\mathrm{a}) \mathrm{How}\) 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 \(\mathrm{C}_{8} \mathrm{H}_{18}(I)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(I)\) by the same amount?
Step-by-Step Solution
Verified Answer
(a) 2664 J of heat is needed for octane. (b) More heat is required for 1 mol of octane than for water.
1Step 1: Understand the formula for heat energy
To find the amount of heat required to change the temperature of a substance, we use the formula: \[ q = m \cdot c \cdot \Delta T \] where \( q \) is the heat energy (in Joules), \( m \) is the mass (in grams), \( c \) is the specific heat capacity (in J/g-K), and \( \Delta T \) is the change in temperature (in Kelvin or Celsius).
2Step 2: Calculate the change in temperature
The change in temperature \( \Delta T \) is calculated by subtracting the initial temperature from the final temperature: \[ \Delta T = \text{final temperature} - \text{initial temperature} = 25.0^{\circ}C - 10.0^{\circ}C = 15.0^{\circ}C \]
3Step 3: Insert values to find heat required for octane
Using the formula from Step 1 and the values given, \( m = 80.0 \) g, \( c = 2.22 \text{ J/g-K} \), and \( \Delta T = 15.0^{\circ}C \), we can calculate: \[ q = 80.0 \cdot 2.22 \cdot 15.0 = 2664 \text{ J} \] Thus, 2664 Joules of heat are required to raise the temperature of 80.0 g of octane from 10.0°C to 25.0°C.
4Step 4: Understanding heat requirement for 1 mol substances
The heat required to increase the temperature of 1 mol of a substance depends on the molar mass and specific heat. Molar mass of \( \mathrm{C}_{8} \mathrm{H}_{18} \) is approximately 114 g/mol. For \( \mathrm{H}_{2} \mathrm{O} \), molar mass is approximately 18 g/mol with specific heat approximately 4.18 J/g-K.
5Step 5: Compare heat for 1 mol octane and water
For 1 mol of \( \mathrm{C}_{8} \mathrm{H}_{18} \), heat is: \[ q = 114 \cdot 2.22 \cdot \Delta T \]. For 1 mol of \( \mathrm{H}_{2} \mathrm{O} \), heat is: \[ q = 18 \cdot 4.18 \cdot \Delta T \]. Calculate: \( 114 \cdot 2.22 = 253.08 \) and \( 18 \cdot 4.18 = 75.24 \). \( 253.08 > 75.24 \), so more heat is needed for octane.
Key Concepts
Heat Energy CalculationTemperature ChangeMolar Mass Comparison
Heat Energy Calculation
To calculate the heat energy required to change the temperature of a substance, we turn to the formula:
For instance, in the case of octane, if you have 80 g of the substance and you are raising the temperature from 10°C to 25°C, the change in temperature \( \Delta T \) would be 15°C. Substituting these values into the formula gives:
- \( q = m \cdot c \cdot \Delta T \)
For instance, in the case of octane, if you have 80 g of the substance and you are raising the temperature from 10°C to 25°C, the change in temperature \( \Delta T \) would be 15°C. Substituting these values into the formula gives:
- \( q = 80.0 \times 2.22 \times 15.0 = 2664 \text{ J} \)
Temperature Change
Temperature change, denoted by \( \Delta T \), is a crucial component in calculating heat energy. It is computed by subtracting the initial temperature from the final temperature:
- \( \Delta T = \text{final temperature} - \text{initial temperature} \)
Molar Mass Comparison
When comparing the heat requirement of different substances, molar mass becomes invaluable. Different substances have different masses in moles, which influences the heat needed to achieve a temperature change. The concept lies in their respective formulas:
When you calculate the energy needed to change the temperature of one mole of each substance by the same amount, you see that:
- For a mole of a substance, \( q = \text{Molar Mass} \times \text{Specific Heat} \times \Delta T \)
When you calculate the energy needed to change the temperature of one mole of each substance by the same amount, you see that:
- Octane: \( 114 \times 2.22 = 253.08 \)
- Water: \( 18 \times 4.18 = 75.24 \)
Other exercises in this chapter
Problem 48
Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(I),\) to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) $$ \mathrm{C}_{6}
View solution Problem 49
(a) Derive an equation to convert the specific heat of a pure substance to its molar heat capacity. (b) The specific heat of aluminum is \(0.9 \mathrm{~J} /(\ma
View solution Problem 55
When an 18.6 -g sample of solid potassium hydroxide dissolves in \(200.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18), the temperature rises
View solution Problem 56
(a) When an \(8.50-\mathrm{g}\) sample of solid ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)\right)\) dissolves in \(120.0 \mathrm{~g}\) of water
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