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TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 56
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 Heat
1. The degree of hotness of a body 2. Base Quantity. 3. Unit: Kelvin (K) @ degree celcius (C) 4. Can measured by using thermometer.
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 Convection Radiation
The vibration of molecules or
atoms from warmer to cooler
areas.
Heat is transferred from one place to
another place by the actual motion of a
hot fluid
Heat energy is transferred by electromagnetic radiation
Conduction occurs in a solid
material.
Convection dominant form of heat
transfer in liquids and gasses.
No transmission medium required.
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
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
Examples: heat transferred by x-rays, radio waves, gamma rays and ultraviolet.
TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 57
they are used to transfer energy
from the stove to the food in pots
and pans.
metal tubes to the atmosphere.
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
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
Q = mc m = mass (kg) c = specific heat capacity (J/kgoC)
= temperature change (oC)
TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 58
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/kg
oC. 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 27C to 37C is 2.64kJ. The mass of the marble is 0.25kg. Calculate the specific heat capacity of the marble.
Q = mc
c = mc
Q
= )2737)(25.0(
)1064.2( 3
= 1101056 CJkg
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 = 452 110 CJkg
Q = mc = (500x10-3) x 452 x (100-20) = 18080Joule
TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 59
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.
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
Q lost = Q gained Heat Lost = Heat Gained
mhot chot (thot - tmixed) = mcold ccold (tmixed - tcold)
TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 60
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 =0.23k110 CJkg ]
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
mcoffeeccoffeecoffee = 747.5
ccoffee = )3)(10 220(
5.7473-
ccoffee 6.1132110 CJkg
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 = 452 110 CJkg ]
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
= 452) x 3(
9000
C064.6
Final temperature = (310C+ C064.6 ) = 37.640C
TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 61
Tutorial Heat and Temperature 1. A tank holding 8 kg of water at 28C 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-1C-1)
2. A piece of metal of mass 0.5 kg is heated to 100C in boiling water. It is then transferred into a well-insulated beaker containing 1.5 kg of water at 27C. If the final steady temperature of the water in the beaker is 32C, 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-1C-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-1C-1 and g = 9.81 ms-2).
4. If 2.5 kg of hot water at 100C is added to 10 kg of cold water at 28C 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 20C of water in in it needs 200kJ energy to heat the water until 80C. Calculate the mass of the water. (Specific heat capacity of water is 4200 J kg-1C-1)
6. A 1.5 kW water heater is used to boil 2.5 kg of 20C water. Calculate the time required to boil the water temperature reaches 100C. (Specific heat capacity of water is 4200Jkg-1C-1)
7. A block of iron of mass 2 kg at temperature 30C 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-1C-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
-1C-1 and cwater=4200 J kg-1C-1]
Answer: 1) 41.4C 2) 926.47 J/kgC 3) 2.46C 4) 42.4C 5) 0.79 kg 6) 560 s 7) 36.6C 8) 34.88 0C 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)
TEMPERATURE AND HEAT BB101- ENGINEERING SCIENCE
UNIT SAINS JMSK PUO/DIS 2012 Page 62
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
Temperature and Heat Tutorial) and let them to
spend several minutes to think about and writing
down the answers.
Set clear expectations regarding the focus
of the thinking and sharing to be done.
Put students in pairs to share and clarify
their ideas and understanding.
Monitor students dialogue by circulating
and listening.
Formulate thoughts and ideas, writing them
down as necessary to prepare for sharing with a
partner.
Practise good active listening skills when
working in pairs, using techniques such as
paraphrasing what the other has said, asking for
clarification, and orally clarifying their own
ideas and answers.
After
Call upon some pairs to share their learning
and ideas with the whole class.
Possibly extend the Think/Pair/Share with a
further partner trade, where students swap
partners and exchange ideas/ answers again.
Pinpoint any information that is still unclear after
the pair discussion, and ask the class and
teacher for clarification.