Prof. Marlon Flores Sacedon Quantity of Heat Prof. Marlon Flores Sacedon Department of Mathematics and Physics College of Arts and Sciences Visayas State University, Visca Baybay City, Leyte, Phiippines

Quantity of Heat When you put a cold spoon into a cup of hot coffee, the spoon warms up and the coffee cools down as they approach thermal equilibrium. The interaction that causes these temperature changes is fundamentally a transfer of energy from one substance to another. Energy transfer that takes place solely because of a temperature difference is called heat flow or heat transfer, and energy transferred in this way is called Quantity of Heat (Q).

MFS Quantity of Heat Where: Q = quantity of heat (J) m = mass (kg) T = change in temperature (K) c = specific heat capacity (J/kg. K ) 1 cal = 4.186 J 1 kcal = 1000 cal = 4186 J 1 Btu = 778 ft.lb = 252 cal = 1055 J Specific heat capacity Specific heat capacity of water: 4190 J/kg.K 1 cal/g. Co 1 Btu/lb. Fo MFS

Specific Heat Capacity, c Approximate specific and molar heat capacities (constant pressure) Substance Specific Heat Capacity, c (J/Kg.K) M (kg/mol) Molar Heat Capacity, C (J/mol.K) Aluminum 910 0.0270 24.6 Beryllium 1970 0.00901 17.7 Copper 390 0.0635 24.8 Ethanol 2428 0.0461 111.9 Ethylene glycol 2386 0.0620 148.0 Ice (near 0oC) 2100 0.0180 37.8 Iron 470 0.0559 26.3 Lead 130 0.201 26.9 Marble 879 0.0585 87.9 Mercury 138 27.7 Salt 51.4 Silver 234 0.108 25.3 Water 4190 75.4

Quantity of heat in terms molar heat capacity No. of moles mass Molar mass OR (Quantity of heat in terms molar heat capacity) Molar heat capacity Specific heat capacity Where: dQ = quantity of heat (Cal) n = No. of moles (mols) dT = change in temperature (K) C = Molar heat capacity (J/mol. K ) Molar mass Molar heat capacity of water = 75.4 J/mol.K

Example 1: During a bout with the flu an 80-kg man ran a fever of 2 Example 1: During a bout with the flu an 80-kg man ran a fever of 2.0Co above normal, that is, a body temperature of 39 oC (102.2 0F) instead of the normal 37oC (98.6 oF). Assuming that the human body is mostly water, how much heat is required to raise his temperature by that amount? Ans: 6.7x105J Example 2: An aluminum tea kettle with mass 1.50 kg and containing 1.80 kg of water is placed on a stove. If no heat is lost to the surface surroundings, how much heat be added to raise the temperature from 20oC to 85oC Ans: 5.79X105J

Specific Heat Capacity, c Approximate specific and molar heat capacities (constant pressure) Substance Specific Heat Capacity, c (J/Kg.K) M (kg/mol) Molar Heat Capacity, C (J/mol.K) Aluminum 910 0.0270 24.6 Beryllium 1970 0.00901 17.7 Copper 390 0.0635 24.8 Ethanol 2428 0.0461 111.9 Ethylene glycol 2386 0.0620 148.0 Ice (near 0oC) 2100 0.0180 37.8 Iron 470 0.0559 26.3 Lead 130 0.201 26.9 Marble 879 0.0585 87.9 Mercury 138 27.7 Salt 51.4 Silver 234 0.108 25.3 Water 4190 75.4

Problem Cup Coffee 70 𝒐 𝑪 1) A geologist working in the field drinks her morning coffee out of an aluminum cup. The cup has a mass of 0.120kg and its initially at 20 𝒐 𝑪 when she pours in 0.300kg of coffee initially at 70 𝒐 𝑪 . What is the final temperature after the coffee and the cup attain thermal equilibrium? (Assume that coffee has the same specific heat as water and that there is no heat exchange with the surroundings.) Thermal Equilibrium 𝑇 𝑓𝑖𝑛𝑎𝑙 20 𝒐 𝑪 𝟎.𝟏𝟐𝟎𝒌𝒈 𝑄 𝑐𝑢𝑝 + 𝑄 𝑐𝑜𝑓𝑓𝑒𝑒 =0 𝑚 𝑐𝑢𝑝 𝑐 𝐴𝑙 𝑇 𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑢𝑝 + 𝑚 𝑐𝑜𝑓𝑓𝑒𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟 𝑇 𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑜𝑓𝑓𝑒𝑒 =0 𝑇 𝑓𝑖𝑛𝑎𝑙 = 𝑚 𝑐𝑢𝑝 𝑐 𝐴𝑙 𝑇 𝑐𝑢𝑝 + 𝑚 𝑐𝑜𝑓𝑓𝑒𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟 𝑇 𝑐𝑜𝑓𝑓𝑒𝑒 𝑚 𝑐𝑢𝑝 𝑐 𝐴𝑙 + 𝑚 𝑐𝑜𝑓𝑓𝑒𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟 =66 𝒐 𝑪

Phase Changes Plasma FORMS OF SUBSTANCE

Phase Changes Freezing Freezing - the substance changes from a liquid to a solid

Phase Changes Melting- the substance changes from a solid to a liquid Melting (Fusion) Freezing Melting- the substance changes from a solid to a liquid

Phase Changes Condensation Melting (Fusion) Freezing Condensation- the substance changes from a gas to a liquid

Phase Changes Vaporization Condensation Melting (Fusion) Freezing Vaporization- the substance changes from a liquid to a gas

Phase Changes Vaporization Condensation Sublimation Melting (Fusion) Freezing Sublimation- the substance changes directly from a solid to a gas without going through the liquid phase

Phase Changes Vaporization Condensation Sublimation Deposition Melting (Fusion) Freezing Deposition- the substance changes directly from a gas to a solid without going through the liquid phase

Phase Changes Where: Q = quantity of heat (J) m mass (kg) and Where: Q = quantity of heat (J) m mass (kg) Lf = heat of fusion (J/kg) LV = heat vaporization (J/kg ) Heat of fusion of water: Lf = 3.34x105 J/ kg = 79.6 cal/g = 143 Btu/lb Heat of vaporization of water: Lf = 2.256x106 J/ kg = 539cal/g = 970 Btu/lb

Problem 2) A physics student wants to cool 0.25 kg of Diet Omni-Cola (mostly water), initially at 25 𝒐 𝑪 , by adding ice initially at −20 𝒐 𝑪 . How much ice should she add so that the final temperature will be 0 𝒐 𝑪 with all ice melted if the heat capacity of the container may be neglected? Ans: 69 grams Apply conservation of energy to the system (Cola and ice)

Problem 3) A heavy copper pot of mass 2.0 𝑘𝑔 (including the copper lid) is at a temperature of 150 𝒐 𝑪 . You pour 0.10𝑘𝑔 of water at 25 𝒐 𝑪 into the pot, then quickly close the lid of the pot so that no steam can escape. Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings. a. Final temperature b. Phase change of water

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Assignment 17.57

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