Question

50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in figure. Find the quantity of heat to be supplied to take it from A to B via ADB.

Answer 1

In path ACB,

$\Delta \text{Q}=50\text{cal}=(50\times 4.2)\text{J}$

$=210\text{J}$

$\Delta \text{W}=W_{\text{AC}}+W_{\text{CB}}$

$=50\times {10}^{-3}\times 200\times {10}^{-6}$

$=10\text{J}$

$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}$

$\Rightarrow \Delta \text{U}=\Delta \text{Q}-\Delta \text{W}=(210-10)\text{J}$

$=200\text{J}$

In path ADB, $\Delta \text{Q}=?$

$\Delta \text{U}=200\text{J}$

(Internal energy change between two points is always same)

$\Delta \text{W}=W_{\text{AD}}+W_{\text{DB}}=\Delta \text{W}$

$=0+155\times {10}^3\times 200\times {10}^{-6}$

$=31\text{J}$

$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}$

$=200+31\text{J}=231\text{J}$

$=55\text{cal}$.