In a calorimeter when `M_(1)` gram of ice at -10°C (specific heat = `0.5 "cal g"^(-1) C^(-1)` is added to `M_2` gram of water at 50°C, it is found that all the ice melts and the water is at 0°C. The value of latent heat of ice in cal `g^(-1)` is:

Options

A. `(5M_(1))/M_(2) - 50`
B. `(50 M_(2))/M_(1) -5`
C. `(5 M_(2))/M_(1)`
D. `(5M_(2))/M_(1) - 5`

Answer

B

Explanation

heat taken by ice =heat given by water
`m_(1)S_("ice")(10) + m_(1)L_(f) = m_(2)S_(w)(50)`
`m_(1)/2 xx 10 + m + L = m_(2) xx 50`
`5 + L = 50 (m_(2)/m_(1))`
`L (50 m_(2))/m_(1) -5`