Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 16 ∘c. assume no heat loss to the surroundings.

There are a lot of missing information in this problem. I've found a similar problem which is shown in the attached picture. Let's just use the information there that we don't have here.

From the principle of conservation of energy:

Heat of dry ice + Heat of water = 0
Heat of dry ice = - Heat of water
(m of dry ice)ΔH = -(m of water)(Cp)(ΔT)
where Cp for water is 4.187 kJ/kg·°C

Hence,
(m of dry ice)(ΔH°g - ΔH°s) = -(Density*Volume)(Cp)(ΔT)
where the density of water is 1 kg/L and the molar mass of dry ice is 44 g/mol.

Then,
(m of dry ice)(-393.5 - -427.4 kJ/mol)(44 g/mol) = -(1 kg/L*12L)(4.187 kJ/kg·°C)(16 - 88 °C)
Solving for m of dry ice,
Mass = 2.43 g of dry ice

If it releases 5 tons a day and there are 7 days a take that answer and multiply by 5 percent (0.05) to find the amount released in one week. leave a comment if you need more .

Ibelieve it's celsius. hope this .