2 deleted 53 characters in body edited Feb 8 '15 at 14:10 Ján Lalinský 17.9k1515 silver badges4444 bronze badges Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature. How does this make intuitive sense? The formula defines entropy change. Since it defines new word not conceived before and thus devoid of sense, it is hard to imagine as having "intuitive sense". If your question really is why people use this definition and not other, here is one possible view: Carnot came to conclusion that all reversible cycles operating between two temperatures $$T_1< T_2$$ (acquiring heat $$Q_1$$ and $$Q_2$$) have the same efficiency (work divided by heat consumed) $$\frac{\sum W}{Q_{2}} = 1 - \frac{T_1}{T_2}.$$ Today, this is being derived from 1st and 2nd law of thermodynamics in most textbooks on thermodynamics. It is done most easily for ideal gas, but the result is generally valid for any substance. It is at this point where the division by temperatures enters the discussion. Since $$\sum W = Q_1+Q_2$$, it follows $$\frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0.$$ (sum of reduced heats equals zero). This follows from 1st and 2nd law of thermodynamics. General cyclic process has the same effect on the surroundings of the system as many Carnot cycles tesselating the original cycle in the work diagram, each operating with very small amount of heat. Writing Carnot's equation for all of them and summing the terms, adjacent terms cancel each other and we are left with sum over terms that correspond to boundary elements of the curve representing the general cycle only: $$\sum_{s} \frac{Q_s}{T_s}$$ with both isothermic and adiabatic elements $$s$$. We pass from this sum to loop integral in the thermodynamic space of states $$\mathbf X$$: $$\oint_\Gamma \frac{\mathbf J}{T}\cdot d\mathbf X = 0$$ where $$\mathbf J$$ is such function of $$\mathbf X$$ and $$\Gamma$$ that integral over segment of $$\Gamma$$ (let's call it $$\Delta \Gamma$$) $$\int_{\Delta \Gamma} \mathbf J\cdot d\mathbf X$$ is the heat accepted by the system when it changes state along the curve $$\Delta\Gamma$$. The last equation can be expressed in words this way: the line integral of $$\mathbf J/T$$ along closed curve $$\Gamma$$ in space of thermodynamic equilibrium states $$\mathbf X$$ is always zero. It follows that the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \mathbf J\cdot d\mathbf X$$ (equal to heat accepted by the system) depends on the path chosen to connect thermodynamic equilibrium states $$\mathbf X_i$$ and $$\mathbf X_f$$, but the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J}{T}\cdot d\mathbf X$$ does not depend on it; it only depends on those two states. This enables us to define function in the space of equilibrium states $$S(\mathbf X_f) = S(\mathbf X_i) + \int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J \cdot d\mathbf X}{T},$$ where $$S(\mathbf X_i)$$ is value of the function for reference state $$\mathbf X_i$$, chosen by convention. It does not matter which path is chosen to connect $$\mathbf X_i$$ and $$\mathbf X_f$$; value of $$S$$ only depends on the endpoint $$\mathbf X_f$$. This function is called entropy. Unfortunately, there is nothing intuitive about it in thermodynamics; it's just a useful definition. Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature. How does this make intuitive sense? The formula defines entropy change. Since it defines new word not conceived before and thus devoid of sense, it is hard to imagine as having "intuitive sense". If your question really is why people use this definition and not other, here is one possible view: Carnot came to conclusion that all reversible cycles operating between two temperatures $$T_1< T_2$$ (acquiring heat $$Q_1$$ and $$Q_2$$) have the same efficiency (work divided by heat consumed) $$\frac{\sum W}{Q_{2}} = 1 - \frac{T_1}{T_2}.$$ Today, this is being derived from 1st and 2nd law of thermodynamics in most textbooks on thermodynamics. It is done most easily for ideal gas, but the result is generally valid for any substance. It is at this point where the division by temperatures enters the discussion. Since $$\sum W = Q_1+Q_2$$, it follows $$\frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0.$$ (sum of reduced heats equals zero). This follows from 1st and 2nd law of thermodynamics. General cyclic process has the same effect on the surroundings of the system as many Carnot cycles tesselating the original cycle in the work diagram, each operating with very small amount of heat. Writing Carnot's equation for all of them and summing the terms, adjacent terms cancel each other and we are left with sum over terms that correspond to boundary elements of the curve representing the general cycle only: $$\sum_{s} \frac{Q_s}{T_s}$$ with both isothermic and adiabatic elements $$s$$. We pass from this sum to loop integral in the thermodynamic space of states $$\mathbf X$$: $$\oint_\Gamma \frac{\mathbf J}{T}\cdot d\mathbf X = 0$$ where $$\mathbf J$$ is such function of $$\mathbf X$$ and $$\Gamma$$ that integral over segment of $$\Gamma$$ (let's call it $$\Delta \Gamma$$) $$\int_{\Delta \Gamma} \mathbf J\cdot d\mathbf X$$ is the heat accepted by the system when it changes state along the curve $$\Delta\Gamma$$. The last equation can be expressed in words this way: the line integral of $$\mathbf J/T$$ along closed curve $$\Gamma$$ in space of thermodynamic equilibrium states $$\mathbf X$$ is always zero. It follows that the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \mathbf J\cdot d\mathbf X$$ (equal to heat accepted by the system) depends on the path chosen to connect thermodynamic equilibrium states $$\mathbf X_i$$ and $$\mathbf X_f$$, but the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J}{T}\cdot d\mathbf X$$ does not depend on it; it only depends on those two states. This enables us to define function in the space of equilibrium states $$S(\mathbf X_f) = S(\mathbf X_i) + \int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J \cdot d\mathbf X}{T},$$ where $$S(\mathbf X_i)$$ is value of the function for reference state $$\mathbf X_i$$, chosen by convention. It does not matter which path is chosen to connect $$\mathbf X_i$$ and $$\mathbf X_f$$; value of $$S$$ only depends on the endpoint $$\mathbf X_f$$. This function is called entropy. Unfortunately, there is nothing intuitive about it in thermodynamics; it's just a useful definition. Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature. How does this make intuitive sense? The formula defines entropy change. Since it defines new word not conceived before and thus devoid of sense, it is hard to imagine as having "intuitive sense". If your question really is why people use this definition and not other, here is one possible view: Carnot came to conclusion that all reversible cycles operating between two temperatures $$T_1< T_2$$ (acquiring heat $$Q_1$$ and $$Q_2$$) have the same efficiency (work divided by heat consumed) $$\frac{\sum W}{Q_{2}} = 1 - \frac{T_1}{T_2}.$$ Today, this is being derived from 1st and 2nd law of thermodynamics in most textbooks on thermodynamics. It is done most easily for ideal gas, but the result is generally valid for any substance. It is at this point where the division by temperatures enters the discussion. Since $$\sum W = Q_1+Q_2$$, it follows $$\frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0.$$ (sum of reduced heats equals zero). General cyclic process has the same effect on the surroundings of the system as many Carnot cycles tesselating the original cycle in the work diagram, each operating with very small amount of heat. Writing Carnot's equation for all of them and summing the terms, adjacent terms cancel each other and we are left with sum over terms that correspond to boundary elements of the curve representing the general cycle only: $$\sum_{s} \frac{Q_s}{T_s}$$ with both isothermic and adiabatic elements $$s$$. We pass from this sum to loop integral in the thermodynamic space of states $$\mathbf X$$: $$\oint_\Gamma \frac{\mathbf J}{T}\cdot d\mathbf X = 0$$ where $$\mathbf J$$ is such function of $$\mathbf X$$ and $$\Gamma$$ that integral over segment of $$\Gamma$$ (let's call it $$\Delta \Gamma$$) $$\int_{\Delta \Gamma} \mathbf J\cdot d\mathbf X$$ is the heat accepted by the system when it changes state along the curve $$\Delta\Gamma$$. The last equation can be expressed in words this way: the line integral of $$\mathbf J/T$$ along closed curve $$\Gamma$$ in space of thermodynamic equilibrium states $$\mathbf X$$ is always zero. It follows that the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \mathbf J\cdot d\mathbf X$$ (equal to heat accepted by the system) depends on the path chosen to connect thermodynamic equilibrium states $$\mathbf X_i$$ and $$\mathbf X_f$$, but the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J}{T}\cdot d\mathbf X$$ does not depend on it; it only depends on those two states. This enables us to define function in the space of equilibrium states $$S(\mathbf X_f) = S(\mathbf X_i) + \int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J \cdot d\mathbf X}{T},$$ where $$S(\mathbf X_i)$$ is value of the function for reference state $$\mathbf X_i$$, chosen by convention. It does not matter which path is chosen to connect $$\mathbf X_i$$ and $$\mathbf X_f$$; value of $$S$$ only depends on the endpoint $$\mathbf X_f$$. This function is called entropy. Unfortunately, there is nothing intuitive about it in thermodynamics; it's just a useful definition. 1 answered Feb 8 '15 at 11:23 Ján Lalinský 17.9k1515 silver badges4444 bronze badges Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature. How does this make intuitive sense? The formula defines entropy change. Since it defines new word not conceived before and thus devoid of sense, it is hard to imagine as having "intuitive sense". If your question really is why people use this definition and not other, here is one possible view: Carnot came to conclusion that all reversible cycles operating between two temperatures $$T_1< T_2$$ (acquiring heat $$Q_1$$ and $$Q_2$$) have the same efficiency (work divided by heat consumed) $$\frac{\sum W}{Q_{2}} = 1 - \frac{T_1}{T_2}.$$ Today, this is being derived from 1st and 2nd law of thermodynamics in most textbooks on thermodynamics. It is done most easily for ideal gas, but the result is generally valid for any substance. It is at this point where the division by temperatures enters the discussion. Since $$\sum W = Q_1+Q_2$$, it follows $$\frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0.$$ (sum of reduced heats equals zero). This follows from 1st and 2nd law of thermodynamics. General cyclic process has the same effect on the surroundings of the system as many Carnot cycles tesselating the original cycle in the work diagram, each operating with very small amount of heat. Writing Carnot's equation for all of them and summing the terms, adjacent terms cancel each other and we are left with sum over terms that correspond to boundary elements of the curve representing the general cycle only: $$\sum_{s} \frac{Q_s}{T_s}$$ with both isothermic and adiabatic elements $$s$$. We pass from this sum to loop integral in the thermodynamic space of states $$\mathbf X$$: $$\oint_\Gamma \frac{\mathbf J}{T}\cdot d\mathbf X = 0$$ where $$\mathbf J$$ is such function of $$\mathbf X$$ and $$\Gamma$$ that integral over segment of $$\Gamma$$ (let's call it $$\Delta \Gamma$$) $$\int_{\Delta \Gamma} \mathbf J\cdot d\mathbf X$$ is the heat accepted by the system when it changes state along the curve $$\Delta\Gamma$$. The last equation can be expressed in words this way: the line integral of $$\mathbf J/T$$ along closed curve $$\Gamma$$ in space of thermodynamic equilibrium states $$\mathbf X$$ is always zero. It follows that the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \mathbf J\cdot d\mathbf X$$ (equal to heat accepted by the system) depends on the path chosen to connect thermodynamic equilibrium states $$\mathbf X_i$$ and $$\mathbf X_f$$, but the integral $$\int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J}{T}\cdot d\mathbf X$$ does not depend on it; it only depends on those two states. This enables us to define function in the space of equilibrium states $$S(\mathbf X_f) = S(\mathbf X_i) + \int_{\mathbf X_i}^{\mathbf X_f} \frac{\mathbf J \cdot d\mathbf X}{T},$$ where $$S(\mathbf X_i)$$ is value of the function for reference state $$\mathbf X_i$$, chosen by convention. It does not matter which path is chosen to connect $$\mathbf X_i$$ and $$\mathbf X_f$$; value of $$S$$ only depends on the endpoint $$\mathbf X_f$$. This function is called entropy. Unfortunately, there is nothing intuitive about it in thermodynamics; it's just a useful definition.