The entropy of a sample of a certain substance increases by 0.836 J K-1 on adding reversibly 0.3344 J of heat at constant temperature. The temperature of the sample is:

Question

The entropy of a sample of a certain substance increases by 0.836 J K-1 on adding reversibly 0.3344 J of heat at constant temperature. The temperature of the sample is: (A) 2.5 K (B) 0.3 K (C) 0.016 K (D) 0.4 K

Tardigrade Admin · Accepted Answer

Δ S=(Δ H/T); T=(Δ H/Δ S)=(0.3344/0.836)=0.4 K

Q. The entropy of a sample of a certain substance increases by $0.836 J K^{- 1}$ on adding reversibly $0.3344 J$ of heat at constant temperature. The temperature of the sample is:

$2.5 K$

$0.3 K$

$0.016 K$

$0.4 K$

$Δ S = \frac{Δ H}{T}; T = \frac{Δ H}{Δ S} = \frac{0.3344}{0.836} = 0.4 K$

Q. The entropy of a sample of a certain substance increases by 0.836JK−1 on adding reversibly 0.3344J of heat at constant temperature. The temperature of the sample is:

2.5K

0.3K

0.016K

0.4K