Heat Transfer Problems

Calorimeter – an insulated container that does not allow heat to enter or leave

Heat Transfer Problem (Notes p. 117-118) Link to heat transfer video 100.0 g of water at a temperature of 22.4 oC is placed in a calorimeter. A 75.25 g sample of Al is removed from boiling water at 99.3oC and placed in calorimeter. Final T = 32.9 oC Aluminum cylinder, 75.25 g 100.0 g water at 22.4 oC Boiling water bath at 99.3 oC Final Temp of system = 32.9 oC S of water = 4.18 J/g oC

Data Table Variable/Units Aluminum Water (in calorimeter) s (J/goC) m (mass in grams) T f (Final Temp in oC) Ti (Initial Temp in oC) ∆T = (Tf – Ti) x 4.18 75.25 100.0 32.9 32.9 99.3 22.4 10.5 -66.4

Trace Heat Flow 2ND LAW: HEAT FLOWS FROM HOT METAL (99.3 oC) to COLD WATER IN CALORIMETER (22.4 oC)

Heat lost by Metal = Heat Gained by Water in Calorimeter - Qmetal = + QH2O - [sm∆T]metal = + [sm∆T] H2O - [sAl (75.25)(32.9-99.3)] = + [(4.18)(100.0)(32.9-22.4) - sAl (75.25)(-66.4) = (418)(10.5) sAl (4996.6 ) = 4389 sAl = 4389/ (4996.6 ) = 0.87787 = sAl = 0.878 J/ g oC (3 SF)

Data Table, HW 12-5, #1 Variable/Units Metal Water (in calorimeter) s (J/goC) m (mass in grams) T f (Final Temp in oC) Ti (Initial Temp in oC) ∆T = (Tf – Ti) x 4.18 23.8 50.0 32.5 32.5 100.0 24.0 8.5 -68.5 NOTE: Since density of water = 1.00 g/mL, 50.0 mL of water weighs 50.0 g

Heat lost by Metal = Heat Gained by Water in Calorimeter, (12.5, #1) - Qmetal = + QH2O - [sm∆T]metal = + [sm∆T] H2O - [sAl (23.8)(32.5-100.5)] = + [(4.18)(50.0)(32.5-24.0) - sAl (23.8)(-68.5) = (209)(8.5) sAl (1630.3 ) = 1776.5 sAl = 1776.5/ (1630.3 ) = 1.089543 = sAl = 1.1 J/ g oC (2 SF)

Data Table, HW 12.5, #2 Variable/Units Metal Water (in calorimeter) s (J/goC) m (mass in grams) T f (Final Temp in oC) Ti (Initial Temp in oC) ∆T = (Tf – Ti) x 4.18 5.05 10.0 23.83 23.83 100.00 22.00 1.83 -76.17

Heat lost by Metal = Heat Gained by Water in Calorimeter, (12.5, #1) - Qmetal = + QH2O - [sm∆T]metal = + [sm∆T] H2O - [sAl (5.05)(23.83-100.00)] = + [(4.18)(10.0)(23.83-22.0) - sAl (5.05)(-76.17) = (41.8)(1.83) sAl (384.66 ) = 76.494 sAl = 76.494/ (384.66 ) = 1.9886 = sAl = 0.199 J/ g oC (3 SF)

Data Table, 12.5, #3 Variable/Units Metal Water (in calorimeter) s (J/goC) m (mass in grams) T f (Final Temp in oC) Ti (Initial Temp in oC) ∆T = (Tf – Ti) 0.449 4.18 x 4.28 x 103 45 45 1445 22 23 -1400.

Heat lost by Metal = Heat Gained by Water in Calorimeter, (12.5, #3) - Qmetal = + QH2O - [sm∆T]metal = + [sm∆T] H2O - [(0.449) (x)(45-1445)] = + [(4.18)(4280)(45-22) - [(0.449) (x(-1400.) = (17890.4)(23) (628.6) (x) = 411479.2 (x) = 411479.2 / (628.6) = = 654.596 (x) = 650 g (2 SF)

HW 12-5, #4 ANS: C ; The initial temp of the pan is 26.0 oC and the final temperature of the system is 39.0 oC. Since the temp of the pan is increasing it must be gaining heat from the water. The initial temp of the water must have been greater than 39. Choice C is the only one in which the temp. of the water is greater than 39. The heat flow direction is from water to metal.

HW 12-5, #5 ANS = C. The iron and the water have equal masses, the only difference is the specific heat capacity terms (0.449 vs 4.18). Since the s for water is almost a factor 10 greater than s for iron the temp of water will rise much less when it absorbs the heat lost by the iron. Q = sm∆T ; ∆T = Q/ sm; Iron: ∆T = Q/(0.449)(100) Water: ∆T = Q/(4.18)(100) Q’s are =, m are =; Iron ∆T will ≈ factor of 10 greater Iron ∆T = 1/0.449 = 2.22 ; Water ∆T = 1/4.18 = 0.23

HW 12-5, #6 ANS: B ; Since the two substances are both identical (water) the s values are identical and the masses are exactly equal the TF will exactly halfway between the Ti of each.