- +1-530-264-8006
- info@tutorsglobe.com

18,76,764

Questions

Asked

21,311

Experts

9,67,568

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment
## Heat Measurements, Physics tutorial

Concept of Heat:Prescot Joule first illustrated that heat is the form of energy in the experiment in which mechanical work (energy) was transformed in heat. Other scientists also illustrated that heat from fuel, like gasoline in the engine, may be transformed to mechanical energy when engine is utilized to drive carts, trains and airplanes. In the electric power station, heat from fuel is changed to electrical energy. Unit of energy is Joule (J) that is also unit of estimating mechanical energy and electrical energy. There are larger units like kilojoule (KJ) and megajoule (MJ).

Power is the rate of doing work. It is also stated as rate at which heat energy is given out by the source. For instance, heat energy delivered per second by the gas burner is its power. This power is estimated in watts (W). One watt is hence stated as one joule per second (J/s or Js

^{-1}). Other larger units are kilowatt (KW) and megawatt (MW).Heat Capacity:Source of heat will transfer heat energy to another body. Source is generally at the high temperature while other body being heated is at the lower temperature. When source and other body are in contact, rise in temperature occurs in colder body.

Let t

_{1}^{o}C be initial temperature of body and t_{2}^{o}C be final temperature when Q joules of heat has been supplied. Then change in temperature,ΔΘ = (t

_{2}- t_{1})^{0}cAmount of heat in joules which is capable of changing its temperature through 1

^{o}C isQ.j/ΔΘ

This amount of heat to change temperature of body is explained as heat capacity of body. It is generally represented by symbol H.

By definition, heat (thermal) capacity (H) of the body, is quantity heat (Q) in joules needed to change its temperature by one degree (Celsius or one Kelvin).

Therefore H = QJ/ΔΘ

^{0}C Where, ΔΘ = (t_{2}- t_{1})Therefore H = QJ(t

_{2}- t_{1})^{o}CHence unit of heat (thermal capacity is expressed in joules per Celsius or joules per Kelvin (JK

^{-1 }or J^{o}C^{-1}).Values of H for different bodies are not same. They differ from one body to another.

Specific Heat Capacity:If there are different masses of the substance m

_{1}, m_{2}and m_{3}, it will be seen that to raise their temperatures through 1^{o}C each, they will need different quantities of heat energy Q_{1}, Q_{2}and Q_{3}. Experiments have illustrated that quantities of heat (Q) needed to change temperature through 1^{o}C is proportional to corresponding masses (m).Q ∝ m

Though, if we fix mass of substance to 1kg and we transfer different quantities of heat Q to it, there will be different equivalent changes in temperature ΔΘ. Again, it will be found that quantities of heat Q to change temperature of 1kg mass of body will be proportional to corresponding changes in temperature ΔΘ. Therefore we can write

Q ∝ ΔΘ

Joining these two factors, we get

Q ∝ mΔΘ

Q = CmΔΘ

Where C is the constant of proportionality called as specific heat capacity of substance

Hence C = Q/mΔΘ

Specific heat capacity of the substance is thus stated as amount of heat Q (in joules) needed to raise temperature of 1kg mass of substance through unit degree (1

^{o}C or 1^{o}K).Unit of C is J/kg

^{o}C or Jkg^{-10}C^{-1}or Jkg^{-1}K^{-1}. It can also be estimated in cal. g^{-1}k^{-1}. Value of C varies from one substance to another.Simple Method of Mixtures:One body is at high temperature and other is at the low temperature. In simple method of mixtures, observe that at hot and cold substances being mixed without considering container in which they are being mixed. Principle of conservation of heat energy is being seen very closely. Here, very in brief, explain principle of conservation of heat energy.

This principle defines that the heat lost by the hot body is equal to heat gained by cold body in any system given there is no heat exchange between substances involved and their surroundings.

Inclusion of Calorimeter in Method of Mixtures:Consider calorimeter as one of the main players in heat exchange and principle of conservation of heat energy. Here we have the hot body A at the higher temperature than liquid contained in calorimeter. Therefore, liquid and calorimeter are considered as cold body gaining heat from the hot body.

Let the hot body (A) with mass m

_{1}, Specific heat capacity C_{1}its initial temperature t_{1 }and Final temperature of mixture is = t.Then, Heat lost by hot body:

Q

_{1}= m_{1}c_{1}ΔΘ_{1}Q

_{1}= m_{1}c_{1}(t_{1}- t)Again let cold liquid (B) with mass m

_{2}, specific heat capacity C_{2}with initial temperature t_{2}. Final temperature of mixture is t. Heat gained by cold liquid (B) = Q_{2 }Q

_{2}= m_{2}C_{2}ΔΘ_{2 }Q

_{2}= m_{2}C_{2}(t - t_{2})Let the cold calorimeter (C) with mass m

_{3}, specific heat capacity C_{3}, Specific heat capacity = C_{3}Initial temperature t_{3}and Final temperature of mixture t. Heat gained by cold calorimeter C is Q_{3}.Q

_{3}= m_{3}C_{3}ΔΘ_{3}Q

_{3 }= m_{3}C_{3}(t - t_{3})It can be noted here that liquid (B) and calorimeter (C) are both gaining heat. Thus their changes in temperature will be same.

θ

_{2}= Δθ_{3}(t - t

_{2}) = (t - t_{3})Q

_{3}= m_{3}C_{3}(t - t_{3})In most cases, m

_{3}and C_{3}may not be given for you in heat exchange. Rather, the property of container that is calorimeter may be given in form of thermal capacity (H) of calorimeter. Therefore, quantity of heat gained by calorimeter is stated asQ

_{3}= HΔθ_{3}= m_{3}C_{3}Δθ_{3}Q

_{3}= H(t - t_{2})By applying the principle of conversation of heat energy, relation between quantities Q

_{1}, Q_{2}and Q_{3}isHeat lost = Heat gained

Q

_{1}= Q_{2}+ Q_{3}m

_{1}C_{1}(t_{1}- t) = m_{2}C_{2}(t - t_{2}) + m_{3}C_{3}(t - t_{2})Or m

_{1}C_{1 }(t - t_{1}) = m_{2}C_{2}(t - t_{2}) + H (t - t_{2})Latent Heat:Specific Latent Heat of Fusion:A solid comprises of atoms or molecules held in fixed structure by forces of attraction between them. Such atoms or molecules vibrate about the mean position. When heat is thus supplied to solid kinetic energy of vibration increases hence increasing temperature of solid. Heat supplied is estimate by mCΔθ . At melting point, heat given to solid is utilized to overcome forces of attraction between atoms or molecules that keep solid in its rigid form, and then solid melts.

At this point we state specific latent heat of fusion (LF) as quantity of heat needed to change 1kg mass of a solid at the melting point to liquid at same temperature.

Specific Latent Heat of Vaporization:Unlike solids, the liquid has no definite form; it generally takes shape of its container. It molecules move in random manner inside though molecules are close enough to attract each other. Some molecules, that have greatest kinetic energy, are able to escape through surface. They then exist as vapor outside the liquid. This procedure is known as evaporation and it occurs at all temperatures. Though, boiling occurs at the definite temperature, boiling point that depends on external pressure. Water for instance boils at 100

^{o}C and at the pressure 760mmHg. It does so at the lower temperature when external pressure is lower like boiling point of water at top of the mountain is less than 100^{o}C. Boiling takes place throughout whole volume of liquid while evaporation is the surface phenomenon.At this point, we would state specify latent heat of vaporization as quantity of heat needed to change 1kg mass of liquid at boiling point to vapor at same temperature.

Latent Heat and Internal Energy:When the liquid reaches boiling point, energy required to change it to vapor is: (i) energy or work required to separate liquid molecules from the mutual attraction until they are comparatively far apart in gaseous state. (ii) Energy or work required to push back external pressure so that molecules can escape from the liquid. Latent heat of vaporization is utilized in point (i) that is required to change internal energy of liquid while point (ii) is external work done against external pressure. Work done in this case is stated as

W = pΔV

Where, p is external pressure and ΔV is change in volume 1g of water changes to approx 1672cm

^{3 }of steamHence ΔV = (1672 - )cm

^{3}= (1671 X 10

^{-6})cm^{3}Assume external pressure P = 1.013 x 10

^{5}Nm^{-2}The work done (W) = pΔV is hence

= 1.0 pΔV 13 x 10

^{5}N/m^{2}X 1671 X 10^{-6}m^{3}= 169.3J

Latent heat of vaporization per gram of water = 2260J

Therefore internal energy part of latent heat of vaporization

= (2260 - 169.3)J

= 2090.7J

This is much greater than external work done. Latent heat of fusion is approx 340J. So energy required to overcome bonds between molecules in solid state is much less than energy to form the gaseous molecules from the liquid state.

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online physics tutoring. Chat with us or submit request at info@tutorsglobe.com