entropy= q/t but i t tt i temperture is increased then randomness also increases then why in the expression for entropy t is in the denominator, i mean entropy is inversely proportional to "t"????
2007-02-19 04:25:14 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Your formula is incorrect. The change in entropy in a reversible process is equal to dQ/T integrated. Even as T increases, the entropy still increases. For example, if you integrate 1/x, increasing the upper bound of integration will add less and less to your result, but it still increases. Same thing here. Increasing temperature contributes less and less to increased entropy, but entropy still increases.
2007-02-19 04:34:14 · answer #1 · answered by William 2 · 1⤊ 0⤋
Hi. Heat flows from hot to cold. The higher the environment "T" temperature the slower the heat flow. In cases where "T" is higher than ambient the flow is opposite. dS > q/T for a natural change.
dS = q/T for a reversible change.
2007-02-19 12:34:45 · answer #2 · answered by Cirric 7 · 0⤊ 0⤋