Question

Steam at $100^{\circ} C$ is passed into $20 g$ of water acquires a temperature of $80^{\circ} C$, the mass of water presents will be [Take specific heat of water $1 cal g ^{-1} c ^{-1}$ latent heat of steam $=540 g ^{-1}$ ]

• A. 24 g
• B. 31.5 g
• C. 42.5 g
• D. 22.5 g

Answer 1

Answer: (D)

22.5 g

Answer 2

(d): Here,
Specific heat of water, $s_w = 1 cal g^{-1} \circ C^{-1}$
Latent heat of steam, $L_s =540 cal g^{-1}$
Heat lost by m g of steam at 100$^{\circ}$C to change into
water at 80$^{\circ}$C is
$Q_1 \, \, \, \, \, \, \, \, = m L_s + m s_w \Delta T_ w$
$\, \, \, \, \, \, \, \, \, \, \, \, \, = m \times 540 + m\times 1\times (100-80)$
$= \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 540 m +20 m = 560 m$
Heat gained by 20 g of water to change its
temperature from 10$^{\circ}$C to 80$^{\circ}$C is
$Q_2 = m_w s_w \delta T_w = 20 \times 1 \times (80 -10) =1400$
According to principle of calorimetry
$Q_1= Q_2$
$\therefore \, \, \, 560 m = 1400 \, \, \, or \, \, \, m= 205 g$
Total mass of water present
$\, \, \, \, \, \, \, \, = (20 + m ) g = (20 +205 ) g = 22.5 g$