Question

50kg of ice at (-60°C) is dropped into the water at ood. How much grams of water freeze​

Answer 1

$\large\star{\boxed{\sf{\red{ Answer \ :- }}}}$

Assuming that you forgot to put a little minus sign in front of the 6 C, let’s, therefore, assume that you meant -6. Assume the SH of ice is 0.5 and that of water 1.0 cal/gm. The heat lost by the water equals that gained as manifested by the increase in temperature of the ice. But the end outcome will be that the temperature of the modified mass of ice will equal that of the water at equilibrium.

And then you’ll discover that the calculation is impossible because:

You haven’t specified the amount of water, notwithstanding the fact that ice does not exist at 6 C at up to a few hundred atmospheres pressure.

You need to allow for the fact that ice floats in water and its immersion is constantly varying with time resulting in changes in the relative amount in contact with the water.

You need to factor in the amount of water movement that directly affects the rate of change of shape and immersion, including whether the container is truly thermally isolated to maintain a constant temperature of the liquid water, or even as converted ice as the cold energy of the 50 g lump of ice gets transferred to the liquid water causing some or all to freeze.