BB101 Engineering Science Chapter 6 Temperature and Heat

TEMPERATURE AND HEATBB101- ENGINEERING SCIENCE 6.0 TEMPERATURE AND HEAT At the end of this chapter, students should be able to       define temperature and heat describe the process of heat transfer define heat capacity, specific heat capacity and state its unit. apply the concept and formula in solving problems on heat capacity and specific heat capacity calculate heat energy transferred between two objects at different temperature determine temperature at thermal equilibrium 6.1 Definition of temperature and heat Temperature, T Temperature is measure of the degree of hotness of a body. A hot body has a high temperature whereas a cold body has a low temperature. Heat, Q Heat is a form of energy being transferred from a hot body to a cold body. The total amount of heat in the body depends on the mass, material and temperature of the body. Differences between temperature and heat Temperature 1. The degree of hotness of a body 2. Base Quantity. 3. Unit: Kelvin (K) @ degree celcius (°C) 4. Can measured by using thermometer. Heat 1. A form of energy. 2. Derived quantity. 3. Unit: Joule (J) 4. No specific measuring equipment 6.2 Describe the Process Of Heat Transfer Heat transferred from one place or body to another in three different ways. The three methods are Conduction, Convection and Radiation. Conduction The vibration of molecules or atoms from warmer to cooler areas. Conduction occurs in a solid material. Examples: putting your hand on a stove burner. The amount of energy transferred depends on how conductive the material is. Metals are good conductors, so Convection Heat is transferred from one place to another place by the actual motion of a hot fluid Convection dominant form of heat transfer in liquids and gasses. Examples: The cooling system of a car, in which water is circulated between the hot engine block and the radiator. In the radiator, the convected heat is conducted through thin-walled UNIT SAINS JMSK PUO/DIS 2012 Radiation Heat energy is transferred by electromagnetic radiation No transmission required. medium Examples: heat transferred by xrays, radio waves, gamma rays and ultraviolet. Page 56 TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE they are used to transfer energy metal tubes to the atmosphere. from the stove to the food in pots and pans. 6.3 Definition of Heat Capacity (Q) and Specific Heat Capacity© Heat Capacity, Q The amount of heat energy (Q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (c) multiplied by the change in temperature (final temperature - initial temperature) Q = m x c x (Tf - Ti) = mc Q = mc m = mass (kg) c = specific heat capacity (J/kgoC)  = temperature change (oC) Specific Heat capacity, c  Specific Heat Capacity (c) of a substance is the amount of heat required to raise the temperature of 1kg of the substance by 1oC (or by 1 K).  The units of specific heat capacity are J oC-1 kg-1 or J K-1 kg-1 UNIT SAINS JMSK PUO/DIS 2012 Page 57 TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE 6.4 Calculate Heat Capacity, Specific Heat Capacity and Energy transfer Example 1 How much heat does 25 g of aluminium give off as it cools from 100 oC to 20 oC?. Given, c aluminium = 880 J/kgoC. Solution: Q = mc = 0.025 kg x 880 J/kgoC x (100 oC - 20 oC) = 1.8 kJ Example 2 The amount of heat needed to increase the temperature of a piece of marble from 27°C to 37°C is 2.64kJ. The mass of the marble is 0.25kg. Calculate the specific heat capacity of the marble. Q = mc Q mc (2.64  10 3 ) = (0.25)(37  27) = 1056 Jkg 10C 1 c= Example 3 42 kJ heat is used to raise water temperature from 20oC to 30oC. Determine the mass of the water. For water, c water = 4.2 kJ/kgoC. Solution: Q = mc 42000 J = m x 4.2 kJ/kgoC x (30-20) oC m = 42000 /42000 m = 1 kg Example 4 An iron spoon of mass 500 g is heated from 200C to 1000C. Calculate the heat absorbed by the iron spoon. Given, c iron = 452 Jkg 10C 1 Q = mc = (500x10-3) x 452 x (100-20) = 18080Joule UNIT SAINS JMSK PUO/DIS 2012 Page 58 TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE 6.5 Determine temperature at thermal equilibrium  Thermal Condition: A condition where two objects in thermal contact has no net transfer of heat energy between each other.  Two bodies are said to be in thermal equilibrium when: a) they are at same temperature b) the net rate of heat flow between them is zero or there is no net heat flow between them  When two substances having different temperatures are introduced or kept together, heat energy flows from a substance at higher temperature to a substance at lower temperature. Also, heat continues to be transferred till their temperatures are equalized. mhot Q lost = Q gained Heat Lost = Heat Gained × chot × (thot - tmixed) = mcold × ccold × (tmixed - tcold) Example 1 In preparing tea, 600 g of water at 90 oC is poured into a 200 g pot at 20 oC. What is the final temperature of the water? Given, c pot = 0.84 kJ / kg.oC. Solution: Heat lost by water = Heat gained by pot (mc) water = (mc) pot (0.6 kg)(4.2 kJ / kg.oC)(90 oC - T) = (0.2 kg)(0.84 kJ / kg.oC)(T - 20 oC) (226.8 - 2.52 T) kJ = (0.168 T - 3.36) kJ 230.16 = 2.688 T T = 230.16 2.688 T = 85.6 oC UNIT SAINS JMSK PUO/DIS 2012 Page 59 TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE Example 2 A silver spoon of mass 50.0 g is at a temperature of 23 0C. This spoon is used to stir coffee which is at temperature of 91 0C. The mass of the coffee is 220 g. The final temperature reached by the spoon and coffee after stirring is 88 0C. Calculate a) Heat absorbed by the spoon b) The specific heat capacity of the coffee. [Given, c spoon =0.23k Jkg 10C 1 ] Solution: a) Heat absorbed by the spoon, Q =mc = (50 × 10-3)( 0.23 × 103)(88 – 23) = 747.5 Joule b) Heat released by the coffee = Heat absorbed by the spoon mcoffeeccoffeecoffee = 747.5 ccoffee = 747.5 (220 × 10 -3 )(3) ccoffee  1132.6 Jkg 10C 1 Example 3 A block of iron of mass 3 kg at temperature 310C is heated with an electric heater rated 100 W for 1.5 minutes. Find the rise in temperature and the final temperature of the block of iron. [c iron = 452 Jkg 10C 1 ] Solution: Heat supplied by heater, Q = Power x time = (100 x 1.5 x 60) =9000 Joule Heat received by iron black = 9000 J 9000 = 3 x 452 x  9000 (3 x 452)   6.64 0 C = Final temperature = (310C+ 6.64 0 C ) = 37.640C UNIT SAINS JMSK PUO/DIS 2012 Page 60 TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE Tutorial Heat and Temperature 1. A tank holding 8 kg of water at 28°C is heated by 1.5 kW electric immersion heater for a period of 5 minutes. If the heat lost to the surroundings can be neglected, find the final temperature of the water. (Specific heat capacity of water is 4200 Jkg-1°C-1) 2. A piece of metal of mass 0.5 kg is heated to 100°C in boiling water. It is then transferred into a well-insulated beaker containing 1.5 kg of water at 27°C. If the final steady temperature of the water in the beaker is 32°C, what is the specific heat capacity of the metal? Assume that there is no loss of heat to the surroundings and ignore the heat transferred to the beaker. (Specific heat capacity of water is 4200 J kg-1°C-1) 3. A piece of lead of mass 2 kg is dropped from the top floor of a building 32.5 m high. If its initial gravitational potential energy is converted totally into thermal energy, find the temperature rise of the lead on hitting the ground. (Specific heat capacity pf lead = 130 J kg-1°C-1 and g = 9.81 ms-2). 4. If 2.5 kg of hot water at 100°C is added to 10 kg of cold water at 28°C and stirred well. What is the final temperature of mixture? (Neglect the heat absorbed by container and the heat lost by the surroundings.) 5. A kettle with 20°C of water in in it needs 200kJ energy to heat the water until 80°C. Calculate the mass of the water. (Specific heat capacity of water is 4200 J kg-1°C-1) 6. A 1.5 kW water heater is used to boil 2.5 kg of 20°C water. Calculate the time required to boil the water temperature reaches 100°C. (Specific heat capacity of water is 4200Jkg1 °C-1) 7. A block of iron of mass 2 kg at temperature 30°C is heated with an electric heater rated 100 W for 1 minute. Find the rise in temperature of the bock of iron.(Specific heat capacity of iron is 452 J Kg-1°C-1) 8. 600 g of copper at temperature of 1150C IS PUT IN 300 g of water with initial temperaure 200C. If there is no heat exchange with the outside, find the final temperature. [Given ccopper = 390J kg-1°C-1 and cwater=4200 J kg-1°C-1] Answer: 1) 41.4°C 2) 926.47 J/kg°C 3) 2.46°C 8) 34.88 0C 4) 42.4°C 5) 0.79 kg 6) 560 s 7) 36.6°C Minimum requirement assessment task for this topic: 1 Theory Test & 1 Lab work & End of Chapter 2 Specification of Theory Test: CLO1 - C1, Specification of lab work: CLO2 – (C2, P1), Specification of End of Chapter: CLO3 – (C2, A1) COURSE LEARNING OUTCOME (CLO) Upon completion of this topic, students should be able to: 1. Identify the basic concept of temperature and heat, (C1) 2. Apply concept of temperature and heat to prove related physics principles. (C2,P1) 3. Apply the concept of temperature and heat in real basic engineering problems. (C2, A1) Compliance to PLO : PLO1 , LD1 (Knowledge) – Theory Test 2 PLO2, LD2 (Practical Skill) – Experiment 4 PLO4, LD4 (Critical Thinking & Problem Solving Skills) UNIT SAINS JMSK PUO/DIS 2012 Page 61 TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE In this strategy, students individually consider an issue or problem and then discuss their ideas with a partner. Purpose: Encourage students to think about a question, issue, or reading, and then refine their understanding through discussion with a partner. What teachers do What students do During •Give students several questions (such as Formulate thoughts and ideas, writing them Temperature and Heat Tutorial) and let them to down as necessary to prepare for sharing with a spend several minutes to think about and writing partner. down the answers. • Practise good active listening skills when • Set clear expectations regarding the focus working in pairs, using techniques such as of the thinking and sharing to be done. paraphrasing what the other has said, asking for • Put students in pairs to share and clarify clarification, and orally clarifying their own their ideas and understanding. ideas and answers. • Monitor students’ dialogue by circulating and listening. After • Call upon some pairs to share their learning and ideas with the whole class. Pinpoint any information that is still unclear after • Possibly extend the Think/Pair/Share with a the pair discussion, and ask the class and further partner trade, where students swap teacher for clarification. partners and exchange ideas/ answers again. UNIT SAINS JMSK PUO/DIS 2012 Page 62