Bounty Ended with 25 reputation awarded by Community♦ occurred Feb 16 '15 at 10:13 3 added 88 characters in body edited Feb 8 '15 at 17:49 Inquisitive 1,392513 You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$ In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group. Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. Try to think of the change in entropy as the noticeably of changes in riotous behavior. You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$ In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group. Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$ In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group. Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. Try to think of the change in entropy as the noticeably of changes in riotous behavior. 2 added 512 characters in body edited Feb 8 '15 at 17:41 Inquisitive 1,392513 You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$ In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group. Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$ You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$ In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group. Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. 1 answered Feb 8 '15 at 17:34 Inquisitive 1,392513 You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $$\Delta S_1=\frac{\Delta Q}{T_{lower}}$$ and $$\Delta S_2=\frac{\Delta Q}{T_{higher}}$$ Assume the $$\Delta Q$$ is the same in each case. The denominator controls the "largeness" of the $$\Delta S$$. Therefore, $$\Delta S_1 > \Delta S_2$$