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

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

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