Conservation of energy PE = 1 j PE = 0 j (arbitrary) 1.0 m Work (w) = 1.O N ( 0.1 kg) X 1.0m = 1.0 j...
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Conservation of energy PE = 1 j PE = 0 j (arbitrary) 1.0 m Work (w) = 1.O N ( 0.1 kg) X 1.0m = 1.0 j trary: assign KE = 0j ervation of energy: PE + KE (1/2mv 2 ) = PE’ + KE’ sed conservation of energy: q + PE + KE = q’ + PE’ + KE’ : Heat at constant pressure: q p = H (enthalpy)
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Transcript of Conservation of energy PE = 1 j PE = 0 j (arbitrary) 1.0 m Work (w) = 1.O N ( 0.1 kg) X 1.0m = 1.0 j...
Conservation of energy
PE = 1 j
PE = 0 j (arbitrary)
1.0 m Work (w) = 1.O N ( 0.1 kg) X 1.0m = 1.0 j
Arbitrary: assign KE = 0jConservation of energy: PE + KE (1/2mv2) = PE’ + KE’Revised conservation of energy: q + PE + KE = q’ + PE’ + KE’NOTE: Heat at constant pressure: qp = H (enthalpy)
Heat calculations
∆H ~ ∆t ∆H ~ m∆H=cm∆t
+q Heat in
-q Heat out
∆t
Heat calculations
Substance ∆HHeat
Specific heat Mass Tfinal Tinitial
Water 22170 j 4.18 j/g-Co 312 g 45oC 28oC
Iron -635 j .450 j/g-Co 52.3 g 4oC 31oC
Ethyl alcohol 660 j 2.46 j/g-Co 79 g 21oC 55oC
Glass 32560 j 0.88 j/g-Co 1.00 kg 58oC 21oC
Aluminum 23392 j 0.900 j/g-Co 358 g 98.3oC 25.7oC
Calculate the answers given the other quantities.Notice the units used. Round-off answers only.
Heat calculations
Use the relation: ∆H = cm∆t to solve the following problems:
1. A student heats 250g of water (4.18 j/g-Co) from 19.7oC to 31.5oC. Calculate the amount of heat added. ANS: 12300 j (3 sig. fig.)
A dietitian burns 4.0g (1 serv or 1 tsp) sugar in a coffee cup calorimeterContaining 565 g of water(cw = 4.18j/g-Co) increasing the temperature from
19.8oC to 46.3oC at constant pressure.
2. Calculate ∆Hw. ANS = 62585 j (62600 j ) ← 0’s not significant)
3. Calculate the Calories (Cal) in 1.0 serving of sugar. 1.0 Cal = 4.18 kj ANS = 14.976076 Cal (15.0 Cal)
4. A student cools 250g of titanium(0.5235 j/g-Co) from 31.5oC to 19.7oC. Calculate the amount of heat involved. ANS: -6.18 j(3 sig. fig.)
Heat calculations1. The student cools 115 g of ethylene glycol (c = 2.42 j/g-Co) at 10oC by removing 3.5 kj of heat. Calculate ∆t and the final temperature. ANS: ∆t = -12.6 Co & tf = -2.6 oC
2. The student heats 43.2 g of silver using 74.1 j of heat increasing the temperature from 25.5oC to 32.8oC. Calculate the specific heat. ANS: 0.235 j/g-Co
378 g of gold (10 troy ounces) at 97.2oC are placed in 138 g of water ( c = 4.18 j/g-Co) at 15.2oC. Heat transfers from the gold to the water until the gold – water mixture is21.6oC. Hg + Hw = Hg’ + Hw’ OR -∆Hg = ∆Hw [Conservation of Energy]
3. Calculate the amount of heat transferred to the water. ANS: 3690 j
4. Calculate the temperature change (∆tm) of the metal (gold). ANS: -75.6 Co
5. Calculate the specific heat capacity (c) of the (metal )gold. ANS: 0.129 j/g-Co
