If 100.0 g of ice at -10.0°C is placed in 200. g of water at 90.0°C in an insulated container, what will be the temperature of the system when equilibrium is established? (Sp. heat of H2O(s) = 2.09 J/g-oC, Sp. heat of H2O(l) = 4.18 J/g-°C, heat of fus. of H2O(s) = 333 J/g.
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
If 100.0 g of ice at -10.0°C is placed in 200. g of water at 90.0°C in an insulated container, what will be the temperature of the system when equilibrium is established? (Sp. heat of H2O(s) = 2.09 J/g-oC, Sp. heat of H2O(l) = 4.18 J/g-°C, heat of fus. of H2O(s) = 333 J/g.
Given : Mass of ice = 100.0 g
Initial temperature of ice = -10.0 oC
Mass of hot water = 200 g
And initial temperature of hot water = 90.0 oC.
Assuming the equilibrium temperature of the system is achieved at a temperature above 0.0 oC i.e melting point temperature of water and is T.
Hence the total amount of heat gained by the ice is given by,
=> Q = mCsΔTs + m ΔHfus + mClΔTl
where m = mass of ice = 100.0 g
Cs = specific heat of ice = 2.09 J/g.oC
ΔTs = change in temperature of ice = 0.0 - (-10.0 ) = 10.0 oC (because ice will melt at 0.0 oC)
Cl = specific heat of liquid water = 4.18 J/g.oC
ΔTl = change in temperature of liquid water = T - 0.0 = T
And ΔHfus = heat of fusion of ice = 333 J/g
Hence substituting the values we get,
=> Q = 100.0 X 2.09 X 10 + 100.0 X 333 + 100.0 X 4.18 X T = 35390 + 418 T
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