Question

Steam at ${100}^{o}C$ is passed into $20g$ of water at ${10}^{o}C$ When water acquires a temperature of ${80}^{o}C$, the mass of water present will be

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

Answer 1

Answer: (D)

$22.5g$

$m\left(g\right)$ steam at ${100}^o\to m\left(g\right)$ water at ${100}^{o}C+540m$ ..... (1)
$m\left(g\right)$ water at ${100}^{o}C\to m\left(g\right)$ water at ${80}^{o}C+\left(m\right)\left(1\right)\left(20\right)$ ..... (2)

(1) + (2)
$m\left(g\right)$ steam at ${100}^{o}C\to m\left(g\right)$ water at ${80}^o+560m\left(cal\right)$ ..... (3)
$20g$ water at ${10}^{o}C+\left(20\right)\left(1\right)70\to 20g$ water at ${80}^{o}C$ ..... (4)

from (3) and (4)
$mix+1400cal\to (20+m)g$ water at ${80}^{o}C+560m\left(cal\right)$
$1400=560m$
$2.5=m$

Total mass of water present
$=(20+2.5)g$
$=22.5g$