Chapter 9: Problem 6
How much heat, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g "C to increase its temperature from \(25^{\circ} \mathrm{C}\) to its melting temperature of \(1535^{\circ} \mathrm{C}\) ?
Short Answer
Expert verified
The iron block requires approximately 50886.75 joules or 12162.70 calories to reach its melting temperature.
Step by step solution
01
Understanding the Problem
We need to calculate the heat required to raise the temperature of a 75.0-g iron block from 25 degrees Celsius to its melting point at 1535 degrees Celsius. We will use the specific heat capacity of iron and the mass of the iron block to calculate the heat in joules and then convert the result to calories.
02
Calculate the Temperature Change
The change in temperature, \(\Delta T\), is the final temperature minus the initial temperature. In our case: \(\Delta T = 1535^\circ \mathrm{C} - 25^\circ \mathrm{C} = 1510^\circ \mathrm{C}\).
03
Calculate the Heat in Joules
The heat (Q) added or removed from a substance can be calculated with the equation \(Q = mc\Delta T\), where m is the mass in grams, c is the specific heat capacity in J/g*C, and \(\Delta T\) is the temperature change in Celsius. Plugging in the given values: \(Q = (75.0\text{ g}) \times (0.449\text{ J/g}^\circ\text{C}) \times (1510^\circ\text{C})\).
04
Compute the Heat Required
After multiplying the mass, specific heat, and temperature change, we get the heat quantity in joules: \(Q = 75.0 \times 0.449 \times 1510 \approx 50886.75 \text{ J}\).
05
Convert the Heat to Calories
1 calorie is equivalent to 4.184 joules. To convert joules to calories, use the conversion factor: \( \text{Calories} = \frac{\text{Joules}}{4.184}\). Thus, \( \text{Calories} = \frac{50886.75}{4.184} \approx 12162.70 \text{ cal}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Specific Heat Capacity
Understanding specific heat capacity is crucial when determining the amount of heat required to change the temperature of a substance. Specific heat capacity, denoted by the symbol 'c', is the amount of heat energy needed to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). Each material has its own unique specific heat capacity, which is a measure of how much heat the material can hold.
For example, iron has a specific heat capacity of 0.449 J/g°C, meaning that for each gram of iron, 0.449 Joules of heat are needed to increase its temperature by one degree Celsius. In the context of the exercise, knowing the specific heat capacity of iron allows us to calculate the total heat needed to change the temperature of a 75.0 g iron block. It is important to note that the specific heat capacity is defined under constant pressure, making it a vital concept in thermodynamics and heating calculations.
For example, iron has a specific heat capacity of 0.449 J/g°C, meaning that for each gram of iron, 0.449 Joules of heat are needed to increase its temperature by one degree Celsius. In the context of the exercise, knowing the specific heat capacity of iron allows us to calculate the total heat needed to change the temperature of a 75.0 g iron block. It is important to note that the specific heat capacity is defined under constant pressure, making it a vital concept in thermodynamics and heating calculations.
Temperature Change
Temperature change, represented by \(\Delta T\), is a straightforward concept that signifies the difference between the final and initial temperatures of a substance. Calculating the temperature change is a necessary step in determining the heat energy involved in a temperature adjustment process.
To find out \(\Delta T\), we subtract the initial temperature from the final temperature. As seen in the given problem, if an iron block's temperature needs to be increased from \(25^\circ \mathrm{C}\) to \(1535^\circ \mathrm{C}\), the temperature change would be \(1510^\circ \mathrm{C}\) (\(1535^\circ \mathrm{C} - 25^\circ \mathrm{C}\)). A larger temperature change requires more heat, while a smaller change requires less. This concept is pivotal in thermal calculations as it helps us understand the proportionality between heat energy and temperature variance within a substance.
To find out \(\Delta T\), we subtract the initial temperature from the final temperature. As seen in the given problem, if an iron block's temperature needs to be increased from \(25^\circ \mathrm{C}\) to \(1535^\circ \mathrm{C}\), the temperature change would be \(1510^\circ \mathrm{C}\) (\(1535^\circ \mathrm{C} - 25^\circ \mathrm{C}\)). A larger temperature change requires more heat, while a smaller change requires less. This concept is pivotal in thermal calculations as it helps us understand the proportionality between heat energy and temperature variance within a substance.
Heat Energy Conversion
Heat energy conversion is an essential aspect of thermodynamics and involves changing one unit of heat energy to another. In chemistry and physics, it is common to convert between joules and calories as they are both units measuring heat energy. A joule is the SI unit while the calorie is another popular unit; one calorie is the amount of heat required to raise the temperature of one gram of water by one degree Celsius.
The relationship between joules and calories is fixed, where 1 calorie equals 4.184 joules. This conversion factor is applied after computing the total heat in joules, as demonstrated in the exercise. By knowing this conversion ratio, the heat required for the temperature change of the iron block from the exercise can be expressed not only in joules but also in calories (\(12162.70 \text{ cal}\)), making the information more accessible depending on the context or preference for unit types.
The relationship between joules and calories is fixed, where 1 calorie equals 4.184 joules. This conversion factor is applied after computing the total heat in joules, as demonstrated in the exercise. By knowing this conversion ratio, the heat required for the temperature change of the iron block from the exercise can be expressed not only in joules but also in calories (\(12162.70 \text{ cal}\)), making the information more accessible depending on the context or preference for unit types.