How much ice at a temperature of - 17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 ∘C?

An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of - 17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg. K, the specific heat for ice to be 2100J/Kg. K, and the heat of fusion for water to be 334000J/kg.

The answer to the above question is

Therefore 0.1133 kg ice at a temperature of - 17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

Explanation:

To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows

ΔH = m*c*ΔT

= 0.205*4190 * (79.9 - 31.0) = 42002.655 J

Therefore fore the ice, we have

Total heat = mi*L + mi*ci*ΔTi = mi*334000 + mi * 2100 * (0 - -17.5) = 42002.655 J

370750*mi = 42002.655 J

or mi = 0.1133 kg

Therefore 0.1133 kg ice at a temperature of - 17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C