Question

$2\,g$ of water condenses when passed through $40\,g$ of water initially at $25^{\circ}C$ . The- condensation of steam raises the temperature of water to $54.3^{\circ}C$ . What is the latent heat of steam?

• A. $540\,\,cal/g$
• B. $536\,\,cal/g$
• C. $270\,\,cal/g$
• D. $480\,\,cal/g$

Answer 1

Answer: (A)

$540\,\,cal/g$

Answer 2

Heat required to raise the temperature of $40\, g$ of water from $25^{\circ} C$ to $54.3^{\circ} C$, is equivalent to sum of heat required to condense the steam.
$\therefore$ Heat required to raise the temperature of water by $t ^{\circ} C$ is
$=m_{1} c \Delta t_{1} \ldots$ (i)
where, $c$ is specific heat of water and $m$ the mass.
Heat required to condense steam
$=m_{2} L+M_{2} c \Delta t_{2} \ldots$ (ii)
Equating Eqs. (i) and (ii), we get
$m_{2} L+m_{2} c \Delta t_{2}=m_{1} c \Delta t_{1}$
Given $m_{2}=2 \,g$
$\Delta t_{2}=(100-54.3)^{\circ} C =45.7^{\circ} C$
$m_{1}=40\, g$
$\Delta t_{1}=(54.3-25)^{\circ} C =19.3^{\circ} C$
$= c = 1\, cal\,g ^{-1}$
$\Rightarrow 2 \times L+2 \times 1 \times 45.7$
$=40 \times 1 \times 29.3$
$\Rightarrow 2 L+91.4=1172$
$\Rightarrow L=540.3\, cal\,g ^{-1}$