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 = 0j Conservation of energy:

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Presentation transcript:

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 = 0j Conservation of energy: PE + KE (1/2mv 2 ) = PE’ + KE’ Revised conservation of energy: q + PE + KE = q’ + PE’ + KE’ NOTE: Heat at constant pressure: q p = H (enthalpy)

Heat calculations ∆H ~ ∆t ∆H ~ m ∆H=cm∆t +q Heat in -q Heat out ∆t

Heat calculations Substance∆H Heat Specific heatMassT final T initial Water22170 j 4.18 j/g-C o 312 g45 o C28 o C Iron-635 j.450 j/g-C o 52.3 g4oC4oC31 o C Ethyl alcohol660 j 2.46 j/g-C o 79 g21 o C55 o C Glass32560 j 0.88 j/g-C o 1.00 kg58 o C21 o C Aluminum23392 j j/g-C o 358 g98.3 o C25.7 o C 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-C o ) from 19.7 o C to 31.5 o C. Calculate the amount of heat added. ANS: j (3 sig. fig.) A dietitian burns 4.0g (1 serv or 1 tsp) sugar in a coffee cup calorimeter Containing 565 g of water(c w = 4.18j/g-C o ) increasing the temperature from 19.8 o C to 46.3 o C at constant pressure. 2. Calculate ∆H w. ANS = 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 = Cal (15.0 Cal) 4. A student cools 250g of titanium( j/g-C o ) from 31.5 o C to 19.7 o C. Calculate the amount of heat involved. ANS: j(3 sig. fig.)

Heat calculations 1. The student cools 115 g of ethylene glycol (c = 2.42 j/g-C o ) at 10 o C by removing 3.5 kj of heat. Calculate ∆t and the final temperature. ANS: ∆t = C o & t f = -2.6 o C 2. The student heats 43.2 g of silver using 74.1 j of heat increasing the temperature from 25.5 o C to 32.8 o C. Calculate the specific heat. ANS: j/g-C o 378 g of gold (10 troy ounces) at 97.2 o C are placed in 138 g of water ( c = 4.18 j/g-C o ) at 15.2 o C. Heat transfers from the gold to the water until the gold – water mixture is 21.6 o C. H g + H w = H g ’ + H w ’ OR -∆H g = ∆H w [Conservation of Energy] 3. Calculate the amount of heat transferred to the water. ANS: 3690 j 4. Calculate the temperature change (∆t m ) of the metal (gold). ANS: C o 5. Calculate the specific heat capacity (c) of the (metal )gold. ANS: j/g-C o