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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :