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

$\begin{array}{l}{\mathrm{Q}}_1=10\times 1\times 10=100\mathrm{cal}\\ {\mathrm{Q}}_2=10\times 0.5[0-(-20\left)\right]+10\times 80\\ =(100+800)\mathrm{cal}=900\mathrm{cal}\end{array}$
As Q₁ < Q₂, so ice will not completely melt and final temperature 0°C.
As heat given by water in cooling upto 0°C is only just sufficient to increase the temperature of ice from -20°C to 0°C hence mixture in equilibrium will consist of 10 gm of ice and 10 gm of water, both at 0°C.