50.25 g of metal that is at a temperature of 10.0 C is placed into 45.0 g H2O which is at 21.5 c, the final temperature of both metal and H2O is 30.3 C. (Given: CH2O = 4.18 J/g C) a. calculate the heat gained by the H2O. b. calculaye the specific heat of the metal c. calculate the density of the hydrogen gas at 30.0 C and 750 mm pressure. (R= 0.0821 L atm/K mol)
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.
50.25 g of metal that is at a temperature of 10.0 C is placed into 45.0 g H2O which is at 21.5 c, the final temperature of both metal and H2O is 30.3 C. (Given: CH2O = 4.18 J/g C)
a. calculate the heat gained by the H2O.
b. calculaye the specific heat of the metal
c. calculate the density of the hydrogen gas at 30.0 C and 750 mm pressure. (R= 0.0821 L atm/K mol)
a) the heat gain by the water is given by
q = mCΔT
where m = mass of water
q = heat
C = specific heat of water = 4.18 J/g.C
ΔT = change in temperature of water = final temperature of water - initial temperature of water = 30.3 - 21.5 = 8.8 C
hence substituting the given values we get
q = 45 X 4.18 X 8.8 = heat gained by water = 1655.28 J
b) Since the heat gained by water is lost by the metal only
Hence the q of metal will be = -q of water
its -ve because metal is losing heat
Hence q of metal = -1655.28
also q of metal = m X C X ΔT
where m = mass of metal
C = specific heat of metal and ΔT = change in temperature of metal = 30.3 - 100 = -69.7 C ( you have mistyped the initial temperature of metal as 10.0 instead of 100. please correct it)
Hence substituting the values we get
-1655.28 = 50.25 X C X (-69.7)
=> C = specific heat of metal = 0.473 J/g C
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