Explanation

(a) Heat required to raise the temperature of 40 g of water from 25$℃$ to 54.3$℃$, is equivalent to sum of heat required to condense the steam.
$\therefore Heat\ required\ to\ raise\ the\ temperature\ of\ water\ by\ t℃\ $is
$= m_{1}c\mathrm{\Delta}t_{1}$ …(i)
Where c is specific heat of water and $m\ $the mass. Heat required to condense steam
$= m_{2}L + m_{2}c\mathrm{\Delta}t_{2}$ …(ii)
Equating eqs. (i) and (ii), we get
$m_{2}L + m_{2}c\mathrm{\Delta}t_{2} = m_{1}c\mathrm{\Delta}t_{1}$
Given, $m_{2} = 2\ g$
$\mathrm{\Delta}t_{2} = (100 - 54.3)℃ = 45.7℃$
$m_{1} = 40\ g$
$\mathrm{\Delta}t_{1} = (54.3 - 25)℃ = 29.3℃$
$c = 1\ calg^{- 1}$
$\Rightarrow 2 \times L + 2 \times 1 \times 45.7 = 40 \times 1 \times 29.3$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2L + 91.4 = 1172$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2L = 1080.6$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = 540.3\ calg^{- 1}$