Ice at `0^@C` is added to 200 g of water initially at `70^@C` in a vacuum flask. When 50 g of ice has been added and has all melted the temperature of the flask and contents is `40^@C`. When a further 80 g of ice has been added and has all melted the temperature of the whole becomes `10^@C`. Find the latent heat of fusion of ice.

Options

A. `80 cal//g`
B. `90 cal//g`
C. `70 cal//g`
D. `540 cal//g`

Answer

B

Explanation

According to principle of calorimetry
`ML_f+MsDeltaT=(msDeltaT)_(water)+(msDeltaT)_(flask)`
`50L_F+50xx1xx(40-0)`
`200xx1xx(70-40)+W(70-40)`
`50L_F+2000=(200+W)30`
`5L_F=400+3W` .(i)
Now the system contains `(200+50)` g of water at `40^@C`, so when further 80 g of ice is added
`80L_F+80xx1xx(10-0)`
`=250xx1xx(40-10)+W(40-10)`
`80L_F=670+3W` .(ii)
solving Eqs. (i) and (ii),
`L_F=90(cal)/(g)` and `W=(50)/(3)g`