Question

$20\text{gm}$ ice at $-10{}^{\circ }\text{C}$ is mixed with $m\text{gm}$ steam at $100{}^{\circ }\text{C}$. The minimum value of $m$ so that finally all ice and steam converts into water at $0{}^{\circ }\text{C}$ is
$($Use: specific heat and latent heat as $C_{ice}=0.5\text{cal/gm}{}^{\circ }\text{C},C_{\text{water}}=1\text{cal/gm}{}^{\circ }\text{C},L_{fusion}$ $=80\text{cal/gm}$ and $L_{vapor}$ $=540\text{cal/gm}$ $)$

• A. $\frac{185}{27}\text{gm}$
• B. $\frac{135}{17}\text{gm}$
• C. $\frac{85}{32}\text{gm}$
• D. $\frac{113}{17}\text{gm}$

Answer 1

The correct option is C $\frac{85}{32}\text{gm}$

For minimum value of $m$, the final temperature of the mixture must be $0{}^{\circ }\text{C}$.
We know that,
Heat lost by steam $=$ Heat gained by ice
$\Rightarrow m{L}_{vapor}+m{C}_{water}\Delta T=m_{ice}{C}_{\text{ice}}\Delta T+m_{ice}{L}_{fusion}$
$\Rightarrow m\times 540+m\times 1\times 100=20\times 0.5\times 10+20\times 80$
$\Rightarrow m=\frac{85}{32}\text{gm}$