If 2 g of ice at $$0^\circ\text{C}$$ is added to 2 g of water at $$60^\circ\text{C}$$, Find the temperature of resultant mixture? What are the resulting contents?

PLACEHOLDER ANSWER - the question has two answers. Melting all the ice needs $$2 \times 80 = 160\text{ cal}$$, but the water can give only $$2 \times 1 \times 60 = 120\text{ cal}$$ in cooling to $$0^\circ\text{C}$$. So only $$\frac{120}{80} = 1.5\text{ g}$$ of ice melts and the mixture stays at $$0^\circ\text{C}$$. It contains 0.5 g of ice and 3.5 g of water.