10 gm of ice at -20°C is dropped into a calorimeter containing l0 gm of water at O°C the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain:

When temperature of ice rises to $0^{0}C$, heat absorbed by it =10$\times 0.5\times (0-(-20\left)\right)=100 cal$.

This amount of heat is rejected by water at $0^{0}C.$

So mass of water frozen = $\frac{100}{80}=1.25 gm$

 Hence mixture in equilibrium will consist of 11.25 gm of ice and 8.75 gm  of water, both at 0°C.