# When $M_{1}$ gram of ice at $-10^{\circ}C$ ( specific heat = $0.5$ cal $g^{-1}^{\circ}C^{-1}$) is added to $M_{2}$ gram of water at $50^{\circ}C$, finally no ice is left and the water is at $0^{\circ}C$. The value of latent heat of ice, in cal $g^{-1}$ is : Option 1) $\frac{5M_{1}}{M_{2}}-50$ Option 2) $\frac{50M_{2}}{M_{1}}$ Option 3) $\frac{5M_{2}}{M_{1}}-5$ Option 4) $\frac{50M_{2}}{M_{1}}-5$

Latent Heat -

Q = mL

- wherein

Q = heat supplied

m = mass of substance

L = latent heat

As we know that

heat taken by ice = heat given by water

$M_{1}$$S _{ice}$ $(10)+M_{1}Lg$ = $M_{2}$ $S _{water}$$(50)$

$\Rightarrow \frac{M_{1}}{2}\times10+M_{1}Lg=M_{2}\times50$

$\Rightarrow 5+Lg=50\frac{M_{2}}{M_{1}}$

$\Rightarrow Lg=\frac{50M_{2}}{M_{1}}-5$

Option 1)

$\frac{5M_{1}}{M_{2}}-50$

Option 2)

$\frac{50M_{2}}{M_{1}}$

Option 3)

$\frac{5M_{2}}{M_{1}}-5$

Option 4)

$\frac{50M_{2}}{M_{1}}-5$

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