Lecture 7: Quantity of Heat PDF

Lecture 7 Quantity of Heat Dr. Marwa A. Abd El Wahaab Department of Engineering Mathematics and Physics Faculty of Engineering, Mansoura University Heat energy (Q) Q = m c ΔT (One phase; solid cs or liquid cL or gaseou...

Lecture 7 Quantity of Heat Dr. Marwa A. Abd El Wahaab Department of Engineering Mathematics and Physics Faculty of Engineering, Mansoura University Heat energy (Q) Q = m c ΔT (One phase; solid cs or liquid cL or gaseous cg state) Lf Q=mL (Phase change) Solid Liquid Lv c : specific heat capacity Liquid Gas (SI unit: J/kg K, British unit: Btu/Ib oF ) L f : latent heat of fusion L v : latent heat of vaporization (SI unit: J/kg, British unit: Btu/Ib) Q=mLf m : mass of the substance that melted Notes Q=mLv m : mass of the substance that vaporized Q (lost) = Q (gained) Specific heat (c) & Latent heat (L) 𝑄 L= 𝑚 Amount of heat energy required to raise Amount of heat energy required to the temperature of a unit of mass of change the phase of a unit of mass that substance by 10 temperature Substance has a higher specific heat capacity  takes a long time to warm or cool Substance has a lower specific heat capacity  takes a short time to warm or cool Specific heat (c) Substances Specific heat Substances Specific heat [J/kg. 0C] [J/kg. 0C] Aluminum 900 Silver 234 Beryllium 1830 Brass 380 Cadmium 230 Wood 1700 Copper 387 Glass 837 Germanium 322 Ice 2090 Gold 129 Marble 860 Iron 488 Alcohol 2400 Lead 128 Mercury 140 Silicon 703 Water 4186  Phase change Lf Lv L f : Latent heat of fusion(melting) = Latent heat of freezing At melting point L v : Latent heat of vaporization = latent of condensation At boiling point  Latent heat of fusion and vaporization for some substances Substance Melting point Latent heat of Boiling point Latent heat of 0C fusion 0C vaporization Helium -269.65 5.23 × 103 -268.93 2.09× 104 Nitrogen -209.97 2.55× 104 -195.81 2.01× 105 Oxygen -218.79 1.38× 104 -182.97 2.13× 105 Ethyl alcohol -114 1.04× 105 78 8.54× 105 Water 0 3.33× 105 100 2.26× 106 Sulfur 119 3.81× 104 444 3.26× 105 Lead 327.3 2.45× 104 1750 8.7× 105 Aluminum 660 3.97× 105 2450 1.14× 107 Silver 960.8 8.82× 104 2193 2.33× 106 Gold 1063 6.44× 104 2660 1.58× 106 Lv > Lf Copper 1083 1.34× 105 1187 5.06× 106  Heating curve of water Water is a good example for illustrating the phase changes as a result of the addition of heat Q = m c ΔT Q α ΔT [No phase change] Q=mL At constant temperature [Phase change]  T 0C 120 0C Q = m Lv Q5 100 0C Boiling point Q4 Q = m c ΔT Q3 Qt= Q1+Q2+Q3+Q4+Q5 Melting point Q = m Lf 0 [0C Q2 Q1 - 20 0C QJ Q1 = mice cice ΔT Q2 = mice Lf Q3= mice cw ΔT Q4 = mice Lv Q5= mice cv ΔT  T 0C 120 0C 3 water + steam 4 100 0C Boiling point 1 ice at 0 0C 2 water at 0 0C 3 water at 100 0C 4 steam at 100 0C Melting point ice + water 1 2 0 [0C ice - 20 0C QJ Absolute zero (0 K) The temperature at which the molecules stop moving. Melting point The temperature at which solid to liquid (phase change) occur freezing point The temperature at which liquid to solid (phase change) occur Boiling point The temperature at which liquid to gas (phase change) occur Condensation point The temperature at which gas to liquid (phase change) occur  The law of conservation of energy Energy can neither be created nor destroyed It can only be transformed from one form to another Q lost = Q gained Has high Has low temperature temperature  Example 3.4 What mass of steam initially at 130 ºC is needed to warm 200 g of water in a 100 g glass container from 20 ºC to 50 º C 130 0C Solution Qs 50 0C Qw +Qg 20 0C ms= ?, Ts= 130 0C, mw= 0.2 kg, mg= 0.1 kg Qlost = Qgained Q steam =Q water + Q glass ms cs Δ T + ms Lv + ms cw Δ T = mw cw Δ T + mg cg Δ T ms ×2100 ×(130-100)+ ms × 2.26× 106 + ms ×4186 ×(100-50)= 0.2×4186×(50-20)+ 0.1×837×(50-20) ms = 10.9 × 10 - 3 kg  MCQ 1. (Final Prog. 2021) A pure substance would freeze or solidify at ………. a. the boiling point b. the condensation point c. the melting point d. zero Fahrenheit 2. (Final Prog. 2021) Unit for latent heat is….. a. Watts per Joule b. Joules per Watt c. Joules per Kilogram d. Pascal per Watt 3. (Final Prog. 2021) The amount of energy required to change the liquid to gas and vice versa without any change in temperature is termed as…… a. Heat Capacity b. Latent Heat of Fusion c. Latent Heat of Vaporization d. Specific Heat Capacity 4. (Final Prep. 2021) The SI unit for the specific heat capacity is…….. a. J. kg -1 °C b. J.kg-1 K-1 c. J K-1 d. J °F-1 5. (Final Prep. 2021) When there is no change of state, upon heating the temperature would….. a. increase b. decrease c. remain constant d. may increase or decrease 6. (Final Prog. 2021) How much a mass of steam at 100 oC will completely melt 3.2 kg of ice at 0 oC? (cw = 4190 J/kg °C, Lv = 2.26 x106 J/kg, Lf = 3.33 x105 J/kg) a.0.397 kg b. 0.47 kg c. 2.54 kg d. 7.34 kg Qlost = Qgain Qsteam = Qice 100 0C mst L v + mst cw ∆T= mice L f mst (2.26 x106 + 4190×100)= 3.2× 3.33 x1050 0C 7. The latent heat of vaporization of a substance (Lv) is….. its latent heat of fusion (Lf). a. smaller than b. the same as c. greater than d. smaller than or equal  MCQ 8. (Final Prog. 2021) How much water at 100°C would be turned into steam at the same temperature by 678 kJ of energy? (Lv = 2.26 x106 J/kg) a. 300 kg b. 0.3 kg 100 0C c. 2.05 kg d. 250 kg Q= mw L v 678 × 103= mw × 2.26 × 106 0 0C 9. (Final Prog. 2021) A 5 kg block of copper (Ccopper = 385 J/kg K) at 20°C is given 46.2 kJ of energy. What will be its final temperature? a.24 oC b. 44 oC c. 297 K d. 277 K Q = mc cc ∆T 46.2 × 103 = 5 × 385× ∆T ∆T = 24 Tf = T1+ ∆T  MCQ 10. (Final Prep. 2021) How many joules of energy are required to change 10 gram of ice at -15 oC to water at 20 oC? (cw = 4190 J/kg °C, cice = 2100 J/kg, Lf = 3.33 x105 J/kg ) a. 4483 J b. 3853 J c. 1466.5 J d. 4483 KJ Q= Q1+ Q2+ Q3 100 0C Q= mice cice ∆T+ mice Lf + mice cw ∆T 20 oC Q= 10×10 - 3 × 2100 ×(0-(-15))+ 10×10 - 3× 3.33 x105 0 0C + 10×10 - 3 × 4190 ×(20-0) - 15 oC  MCQ 11. (Final Prep. 2021) How much water at 50°C is needed to just melt 2.2 kg of ice at 0°C? Take (Cwater= 4200 J/kg. K and Lf = 3.34 ×10 5 J/kg) a. 0.5 kg b. 3.5 kg c. 1.5 kg d. 4.9 kg Q lost (Water)= Q gained (ice) 100 0C mw cw ∆T = mice Lf 50 oC Q2 Q1 mw × 4200 × (50 - 0) = 2.2 × 3.34 ×10 5 0 0C  MCQ  (12, 13, 14, 15) A 150 g of ice in a container at 0°C. (cw = 4190 J/kg °C, cice = 2100 J/kg°C, cst = 2000 J/kg°C, LV = 2.26 x106 J/kg, Lf = 3.33 x105 J/kg) 12. How much energy is needed to melt 100 g of ice? a. 315 J b. 16.65 KJ c. 33.3 KJ d. 49.95 KJ Q = mice L f Q = 100 ×10 - 3 × 3.33 x105 100 0C 13. The time required to convert all the ice into water at 0 0C using a constant power of 1000 W electric heater is…… 0 0C a. 3.15 sec b. 6.285 sec c. 49.95 sec d.56.235 sec Q = mice L f Q = 150 ×10 - 3 × 3.33 x105 Q = 49,950 J Q = power × time 49,950 = 1000 × time  MCQ 14. What is the mass of steam at 100 oC will completely melt all the ice? a. 14.74 g b. 18.6 g c. 22.19 g d. 119.29 g 100 0C Qlost = Qgain Qsteam = Qice mst L v + mst cw ∆T= mice L f mst (2.26 x106 + 4190×100)= 150×10 - 3 x 3.33 x 105 0 0C 15. What would be the final state if the ice received 450 KJ of heat? a. ice b. ice and water c. water d. water and steam Q1= mice L f Q1=150×10 - 3 x 3.33 x 105 = 49,950 J Q2= mice cw ∆T Q2=150×10 - 3 x 4190 × 100= 62,850 J Q1+ Q2 = 112,800 J Q3= mice L v Q3= 150×10 - 3 x 2.26 x 106 = 339,000 J Q1+ Q2 + Q3= 451,800 J  Exercise A 540 g of ice at 0°C is mixed with 540 g of water at 80°C. What is the final temperature of the mixture? ( Lf = 80 cal/g , Cw = 1 cal/ g K) a. 0°C b. 40°C c. 80°C d. Less than 0°C Solution mice = 540 g, Tice = 0 o C, mw = 540 g, Tw = 80o C Water 80 oC Qlost (water)= mw cw Δ T = 540×1×80 = 43200 J Ice 0 oC Ice melted Qgained (ice)= mice Lf = 540 × 80 = 43200 J Qlost = Qgained , Final state is water at Tf = 0 o C, Final temperature is 0 o C Thanks