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)

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 $
$\Rightarrow L=540.3\, cal\,g ^{-1}$