Calorimetry Questions + Answers 1. If it takes 41.8 J to heat 18.69 g of gold from 10.0 to 27°c, what is the specific heat capacity of gold? [0.13 J/g ° C] Given: m = 18.69 g ΔT = Tf – Ti ΔT = 27°c – 10.0°c ΔT = 17°c q = 41.8 J Required: c, specific heat capacity Analysis: q = mcΔT Solution: q = mcΔT !
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= 𝑐 = 𝑐 0.13558241 J/g°c = c 0.13 J/g°c = c Paraphrase: ∴ the specific heat capacity of gold is 0.13 J/g°c 2. A 25.0 g steel ball (c = 0.427 kJ/kg ° C) is quickly transferred from boiling water to a 0.200 kg calorimeter with 50.0 g of water at 25.0° C. If the final temperature is 27.0° C, calculate the heat capacity of the calorimeter. [0.903 kJ/kg ° C] Given: m
s = 25.0 g ΔT
s = Tfs – Tis ΔT = 27°C – 100°C ΔT = -­‐73°C C
s = 0.427 kJ/kg°C mw = 50.0 g ΔTw = Tfw – Tiw ΔT = 27°C – 25°C ΔT = 2°C cw = 4.18 J/g°C mc = 0.200 kg mc = 200 g ΔTc = Tfc – Tic ΔT = 27°C – 25°C Δ T = 2°C Required: c, specific heat capacity Analysis: -­‐[mcΔT] = [mcΔT + mcΔT] Solution: -­‐[mscsΔTs] = [mwcwΔTw + mcccΔTc] -­‐[(25.0 g)(0.427 kJ/ kg°C)( -­‐
73°c)] = [(50.0 g)(4.18 J/g°C)(2°C) + (200 g)(c)( 2°C)] 779.275 J = 418 J + (400 g°C)(c) 361.275 J = (400 g°C)(c) 0.9031875 kJ/kg°C = c 0.903 kJ/kg°C = c Paraphrase: ∴ the specific heat capacity of the calorimeter is 0.903 kJ/kg°C 3. A 237 g piece of molybdenum at 100.0°C is thrust into 244g of water 10.0°C. If the final temperature is 15.3°C, what is the specific heat capacity of molybdenum? [0.270 kJ/kg°C] Given: Required: c, specific heat 20073.9 g°C (c) = 5405.576 m
1 = 237 g capacity J ΔT
1 = Tf1 – Ti1 ΔT = 15.3°C – 100°C Analysis: c = 0.269283796 kJ/kg°C ΔT = -­‐
84.7°C -­‐[mcΔT] = [mcΔT] c = 0.270 kJ/kg°C m2 = 244 g Solution: Paraphrase: ΔT2 = Tf2 – Ti2 -­‐[m1c1ΔT1] = [m2c2ΔT2] ∴ the specific heat capacity ΔT = 15.3°C – 10°C -­‐[(237 g)(c)( -­‐84.7°c)] = of molybdenum is 0.270 ΔT = 5.3°C [(244 g)(4.18 kJ/kg°C c2 = 4.18 J/g°C J/g°C)(5.3°C)] 4. If 100.0 g of zinc (c = 0.377 kJ/kg°C) at 95.0°C is dropped into 60.0g of water contained in a 100.0 g calorimeter (c = 0.419 kJ/kg°C) at 20.0°C, what will the final temperature be? [28.6°C] Given: Required: Tf, final -­‐[(37.7 kJ/°C)(Tf -­‐ 95.0°C)] temperature = [(250.8 J/°C)(Tf -­‐ 20°C) + m1 = 100.0 g (41.9 J/ °C)( Tf -­‐ 20°C)] ΔT1i = 95.0°C Analysis: c1 = 0.377 kJ/kg°C -­‐[mcΔT] = [mcΔT + mcΔT] -­‐[(37.7 kJ/°CTf – 3581.5 kJ)] = [(250.8 J/°CTf – 5016 J) + m2 = 60.0 g Solution: (41.9 J/ °C Tf – 838 J)] ΔT2i = 20°C -­‐[m1c1ΔT1] = [m2c2ΔT2 + c2 = 4.18 J/g°C m3c3ΔT3] -­‐330.4 J/ °C Tf = -­‐9435.5 J Tf = 28.55780872°C m3 = 100.0 g -­‐[(100.0 g)(0.377 kJ/ kg°C)( Tf = 28.6 °C ΔT3i = 20°C Tf -­‐ 95.0°c)] = [(60.0 C3 = 0.419 kJ/kg°C g)(4.18 J/g°C)(Tf -­‐ 20°C) + Paraphrase: (100.0 g)(0.419 kJ/kg°C)( ∴ the final temperature will Tf -­‐ 20°C)] be 28.6°C
5. When 13.0 g of acetylene (C 2 H 2 ) undergoes complete combustion, 65.5 kJ of heat are released. a. How much energy is released when one mole of acetylene is burned? [-­‐
131 kJ/mol] 𝑚𝑎𝑠𝑠
Given: Paraphrase: Analysis: # 𝑜𝑓𝑚𝑜𝑙𝑠 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
m = 13.0 g 131 kJ/mol of energy is ∴ -­‐
q = 65.5 kJ released
𝛥ℎ = 𝑛𝛥ℎr MM = 26 g/mol Paraphrase: Solution: Required: Δhr ∴ the specific heat capacity 𝑚𝑎𝑠𝑠
of gold is 0.13 J/g°c # 𝑜𝑓𝑚𝑜𝑙𝑠 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
13.0 𝑔
# 𝑜𝑓𝑚𝑜𝑙𝑠 =
26 𝑔/𝑚𝑜𝑙
# 𝑜𝑓𝑚𝑜𝑙𝑠 = 0.5 𝑚𝑜𝑙𝑠 𝛥ℎ = 𝑛𝛥ℎr !!
= 𝛥ℎr !
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131 !"#$ = 𝛥ℎr b. Write a balanced chemical equation for the reaction and include the heat value in the equation. 2C2H2 + 5O2  4CO2 + 2H2O + 131 kJ/mol 6. A Bunsen burner flame takes 8 minutes to raise 250 g of water from 20.0 to 100.0°C, when the water is in a beaker of mass 100.0 g (c = 0.670 J/g°C). How long will it take to raise the temperature of 200.0 g of glycerine (c = 2.43 J/g°C) contained in the same beaker from 20.0 to 80.0°C? [3 min] Given: m
1 = 250 g ΔT
1 = Tf1 – Ti1 ΔT = 100°C – 20°C ΔT = 80°C c1 = 4.18 J/g°C m3 = 200.0 g ΔT3 = Tf2 – Ti2 ΔT = 80°C – 20°C = 60°C c = 2.43 J/g°C t = 8 minutes Required: min, time Analysis: q1 = m1c1ΔT1 q3 = m3c3ΔT3 !!
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= ! Solution: q1 = m1c1ΔT1 q1 = (250 g)( 4.18 J/g°C) (80 °C) q1 = 83600 J q3 = m3c3ΔT3 q3 = (200.0 g)( 2.43 J/g°C) (60 °C) q3 = 29160 J !!
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83600 J x = 233280 J/min x = 2.790430622 min x = 3 min Paraphrase: ∴ it will take 3 minutes to raise the temperature of 200.0 g of glycerine contained in the same beaker from 20.0 to 80.0°C