In a container of negligible mass ‘m’ grams of steam at ${100}^{0}C$  is add to 100 gm of water at ${20}^{0}C$  . If no heat is lost to the surrounding at equilibrium, Match the following

	Column – I		Column – II
I)	Mass of steam in the mixture(in gm) : If m = 20 gm	P)	$114.8$
II)	Mass of water in mixture (in gm) if m = 20 gm	Q)	$76.4$
III)	If m = 20 gm find temperature of the mixture (in ${ }^{0}C$)	R)	$5.2$
IV)	If m = 10 gm final temperature of the mixture (in ${ }^{0}C$)	S)	$100$
		T)	$120$

 

Amount of Heart required to convert ${20}^{0}C\mathrm{int}o{100}^{0}C$  of water 

$\begin{array}{l}Qmsd\theta massofsteam\\ =100\times 1\times 300 mL=8,000\\ Q=Cal{m}_s=\frac{8,000}{540}\\ =8000m_s=14.8gm\end{array}$

If  mass of learn is above 14.8 gm the resultant temp =  ${100}^{0}C$
Mass of steam in mixture = 20 – 14.8 = 5.2 gm
Mass of water in mixture = 100 + 14 .8 = 114.8 gm
If m = 10gm         Find temp $\theta ={76.4}^{0}C$