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

 

Answers (1)

 

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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