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Chapter 15: Problem 49
A hot rock ejected from a volcano's lava fountain cools from \(1100^{\circ}\mathrm{C}\) to \(40.0^{\circ} \mathrm{C},\) and its entropy decreases by 950J/K. How much heat transfer occurs from the rock?
Short Answer
Expert verified
The heat transfer is approximately equal to the negative of the product of the entropy change and the average temperature in kelvins.
Step by step solution
01
Understanding the relationship between entropy and heat transfer
The change in entropy (ΔS) of a system is related to the heat transfer (Q) at a constant temperature by the equationΔS = Q/T, where T is the temperature in kelvins. Since the temperature of the rock changes during cooling, we cannot assume a constant temperature. However, we can use an average temperature for approximation if the temperature change is not too drastic.
02
Converting Celsius to Kelvin
First, we need to convert the temperatures from degrees Celsius to kelvins by adding 273.15. The initial temperature (T_i) is 1100°C + 273.15, which equals 1373.15 K. The final temperature (T_f) is 40.0°C + 273.15, which equals 313.15 K.
03
Calculating the average temperature in kelvins
The average temperature (T_avg) of the rock during cooling is the mean of the initial and final temperatures: (T_avg = (T_i + T_f) / 2). This results in (T_avg = (1373.15 K + 313.15 K) / 2).
04
Computing the heat transfer using the entropy change
Now, rearrange the relationship formula to solve for heat transfer, Q = ΔS * T. Since we're using an average temperature, the heat transfer is approximately Q = ΔS * T_avg. Substitute the entropy decrease (-950 J/K) and the average temperature calculated in the previous step to find Q.
05
Final calculation
With the values for ΔS and T_avg, we can calculate the heat transfer: Q = -950 J/K * T_avg. Remember to include the negative sign as the entropy is decreasing, indicating that heat is being lost from the rock.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Entropy
Entropy is a concept in thermodynamics often described as a measure of the disorder or randomness in a system. It's also seen as an indicator of how much energy in a system is unavailable for doing work. When heat is transferred out of an object, like the cooling volcanic rock, the system's entropy typically decreases because it becomes more ordered as it loses energy. In the presented exercise, the entropy of the rock decreased by 950 J/K, which aligns with the notion that the rock's particles are settling into a more orderly state as it cools down.
In mathematical terms, the change in entropy (ΔS) can be thought of in relation to heat transfer (Q) at a constant temperature through the equation ΔS = Q/T, where T is the absolute temperature in kelvins. This relationship is pivotal for understanding heat transfer in thermodynamic processes and is crucial when calculating the amount of heat lost or gained by a system.
Thermodynamics
Thermodynamics is the study of heat, work, and the energy of a system. It's foundational in many areas of science and engineering, particularly when discussing the movement and conversion of heat energy. There are four laws of thermodynamics that govern the behavior of these quantities. The scenario of a cooling volcanic rock directly relates to the first law of thermodynamics, which states that energy cannot be created or destroyed in an isolated system. As the rock cools, the heat energy is not lost to the universe but transferred to the surrounding environment.
This principle allows us to approach the problem by assuming a conservation of energy. The heat that the rock loses as it cools to a lower temperature is the heat transferred to the environment. Hence, by measuring changes in temperature and entropy, we can infer the amount of heat that has been exchanged.
Temperature Conversion
Temperature conversion is necessary when working with thermodynamic equations since these often require using an absolute temperature scale like Kelvin (K). The exercise involves converting Celsius to Kelvin, a common task in thermodynamics that allows for more universal application of equations and comparisons of temperature data.
To convert from Celsius to Kelvin, one adds 273.15 to the Celsius temperature, reflecting the difference in the zero points of the two scales. Celsius is based on the freezing and boiling points of water, while Kelvin is based on absolute zero, the hypothetical lowest possible temperature where all kinetic motion of particles stops. In the volcanic rock problem, temperature conversion is crucial to finding the average temperature during the rock's cooling process, which is then used to calculate the heat transfer.
Kelvin
The Kelvin scale is an absolute temperature scale used primarily in the physical sciences. It's important in thermodynamics because it starts at absolute zero, unlike the Celsius and Fahrenheit scales, which are based on water's properties. Absolute zero, or 0 K, is the point where particles have minimal thermal motion. All thermodynamic calculations involving temperature must use Kelvin to maintain consistency and accuracy.
In the context of our volcanic rock cooling problem, we express temperatures in Kelvin to accurately use the thermodynamic formula for calculating heat transfer. Even though the temperatures are given in Celsius, they're converted to Kelvin before being used to find the average temperature and, subsequently, the amount of heat transferred as the rock cools.
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