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Heat, temperature and internal energy

Heat, temperature and internal energy. When an object is heated, heat energy flows into it and its temperature increases. Hence, heat energy transfer is closely related to temperature change. The transfer of heat to an object would usually lead to an increase of its internal energy .

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Heat, temperature and internal energy

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  1. Heat, temperature and internal energy • When an object is heated, heat energy flows into it and its temperature increases. Hence, heat energy transfer is closely related to temperature change. • The transfer of heat to an object would usually lead to an increase of its internal energy. • Internal energy, which is associated with the microscopic components of the object (i.e. atoms & molecules), includes kinetic energy of translation, rotation and vibration of the atoms/molecules as well as the potential energies of the atoms/molecules.

  2. Heat, temperature and internal energy (cont.) • The internal energy of a system can be changed even when there is no energy transferred by heat. For example: when a gas is compressed, its internal energy increases and so does its temperature. In such a case, the change of the system’s internal energy is caused by the work done to the system. • Generally, when energy is transferred by heat into a substance, its temperature rises and its internal energy increases. However, there are situations in which the increase of internal energy is not accompanied by a rise in temperature. This happens when the substance changes from one physical form to another (e.g. from solid to liquid or from liquid to gas). Such a change is commonly referred to as phase change.

  3. Units of heats Calorie (cal) – One cal is defined as the amount of heat required to raise the temperature of 1 g of water from 14.5oC to 15.5oC. Earlier, calorie was defined as the amount heat required to raise the temperature of 1 g of water by 1oC. However, it was realized subsequently that the amount of heat required to raise 1 g of water by 1oC depends on the temperature of the water. Therefore, the definition was refined. The above definition is sometimes referred to as the “15o calorie”.

  4. Units of heats(con.) British thermal unit (Btu) – One Btu is the amount of heat required to raise the temperature of 1 pound of water from 63oF to 64oF. (1 Btu = 252 cal)

  5. Mechanical equivalent of heat Heat is just a form of energy and can be converted from mechanical energy. Although the idea that heat energy and mechanical energy are connected was first suggested by Benjamin Thomson (1753-1814). It was Joule who established the equivalence of these two forms of energies experimentally. His experimental result was: 1 cal = 4.18 J The equivalence obtained with more precise measurements later is: 1 cal = 4.186 J

  6. Joule’s experiment The falling blocks rotate the paddles, causing the temperature of the water to increase . T 2 m · g · h

  7. 2 m · g · h 2(1.5 kg)( 9.8 m/s2)(3 m) T = = 4.186 Mw (4.186 J/g ·oC)(200g) Example Consider Joule’s apparatus shown in the previous slide. Each of the two mass is 1.5 kg, and the tank is filled with 200 g of water. What is the increase in the temperature of the water after the masses fall through a distant of 3 m? 4.186 Mwater· T = 2 m · g · h = 0.105 oC

  8. Q C = T Heat capacity The heat capacity of a particular sample of a substance is the amount of heat energy required to raise the temperature of that sample by 1oC. Hence, if heat energy Q gives rise to a temperature change of T ofa substance, the heat capacity of the substance is: The unit of heat capacity is obviously J/oC.

  9. Q c = m T Specific heat The specific heat c of a substance is the heat capacity per unit mass, i.e. The unit of specific heat is J/goC.

  10. Energy transferred by heat The energy transferred by heat to a substance from its surrounding can be calculated using the formula: Q = m c T = m c (T2 – T1) This formula is valid only if c is independent of temperature.

  11. In fact, the specific heat of all materials vary somewhat with temperature and also with pressure. When this variation may not be neglected, the amount of heat required to change the temperature of a substance from T1 to T2 should be determined with the following equation:

  12. 4.22 4.21 4.20 4.19 4.18 4.17 c (J/g oC) 0 20 40 60 80 100 T (oC) Specific heat of water as a function of temperature

  13. Measuring specific heat One technique for measuring specific heat involves the following procedure: • Heat a sample to some known temperature. • Place the heated sample in a vessel containing water of known mass and temperature. • Measure the temperature of the water after thermal equilibrium has been reached. This technique is known as calorimetry and the device used for the measurement is called a calorimeter.

  14. mw cw(Tf - Tw) cx = mx (Tx - Tf) Calorimetry Let cx = specific heat of the sample to be determined cw = specific heat of the water mx = mass of the sample mw = mass of the water Tx = initial temperature of the sample Tw = initial temperature of the water (Tw < Tx) Tf = temperature of water and sample at equilibrium Energy gained by water = Energy lost by sample mw cw(Tf - Tw) = - mx cx(Tf - Tx)

  15. Phase change Heat energy is needed to change a substance from solid form to liquid form (melting) or from liquid form to gaseous form (evaporation). Such a change of physical form is called phase change. The energy absorbed by a substance during phase change will not results in any temperature increase. The energy is spent in breaking the bonds between molecules, leading to an increase in potential energy.

  16. The laten heat L of a substance is defined as the amount of heat energy required to change the phase of one unit mass of the substance. Hence, the amount of heat energy needed to change the phase of a substance of mass m is: Q = m L Laten heat (‘hidden heat’ literally) Laten heat of fusion - Laten heat for changing a substance from solid phase to liquid phase. Laten heat of vaporization – laten heat for changing a substance from liquid phase to gaseous phase.

  17. Example How much ice at –20oC must be dropped into 0.25 kg of water, initially at 20oC, in order for the final temperature to be 0oC with the ice all melted? The heat capacity of the container may be neglected. Given that cwater = 4.186 J/g·oC, cice= 2.09 J/g·oC, Lwater = 333 J/g. Let the mass of the ice be m g. Heat required to raise the temperature of the ice from -20oC to 0oC: Q1 = 20 m cic = (20oC) (2.09 J/g·oC)(m g) = 41.8m J Heat required to melt the 0oC ice: Q2 = m Lwater = (m g)(333 J/g) = 333m J

  18. Example (cont) Heat lost by 0.25 kg of water in lowering its temperature from 20oC to 0oC: Q3 = (250 g)(20oC) cwater= (250 g)(20oC)(4.186 J/g·oC) = 20930 J Since Q1 + Q2 = Q3 , therefore 41.8m + 333m = 20930 m = 20930/374.8 = 55.8 g

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