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

Heat, temperature and internal energy (cont.) oThe 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. oGenerally, 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.

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 14.5 o C to 15.5 o C. Earlier, calorie was defined as the amount heat required to raise the temperature of 1 g of water by 1 o C. However, it was realized subsequently that the amount of heat required to raise 1 g of water by 1 o C depends on the temperature of the water. Therefore, the definition was refined. The above definition is sometimes referred to as the “15 o calorie”.

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 63 o F to 64 o F. (1 Btu = 252 cal)

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 ( ). 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 = J

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

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? M water ·  T = 2 m · g · h  T = 2 m · g · h M w 2(1.5 kg)( 9.8 m/s 2 )(3 m) (4.186 J/g · o C)(200g) = = o C

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 1 o C. Hence, if heat energy Q gives rise to a temperature change of  T of a substance, the heat capacity of the substance is: Q TT C = The unit of heat capacity is obviously J/ o C.

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

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 (T 2 – T 1 ) This formula is valid only if c is independent of temperature.

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 T 1 to T 2 should be determined with the following equation:

Specific heat of water as a function of temperature T ( o C) c (J/g  o C)

Measuring specific heat One technique for measuring specific heat involves the following procedure: 1.Heat a sample to some known temperature. 2.Place the heated sample in a vessel containing water of known mass and temperature. 3.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.

Calorimetry Let c x = specific heat of the sample to be determined c w = specific heat of the water m x = mass of the sample m w = mass of the water T x = initial temperature of the sample T w = initial temperature of the water (T w < T x ) T f = temperature of water and sample at equilibrium Energy gained by water = Energy lost by sample m w c w (T f - T w ) = - m x c x (T f - T x ) c x = m w c w (T f - T w ) m x (T x - T f )

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.

Laten heat ( ‘hidden heat’ literally) 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 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.

Example How much ice at –20 o C must be dropped into 0.25 kg of water, initially at 20 o C, in order for the final temperature to be 0 o C with the ice all melted? The heat capacity of the container may be neglected. Given that c water = J/g· o C, c ice = 2.09 J/g· o C, L water = 333 J/g. Let the mass of the ice be m g. Heat required to raise the temperature of the ice from -20 o C to 0 o C: Q 1 = 20 m c ic = (20 o C) (2.09 J/g· o C)(m g) = 41.8m J Heat required to melt the 0 o C ice: Q 2 = m L water = (m g)(333 J/g) = 333m J

Example (cont) Heat lost by 0.25 kg of water in lowering its temperature from 20 o C to 0 o C: Q 3 = (250 g)(20 o C) c water = (250 g)(20 o C)(4.186 J/g· o C) = J Since Q 1 + Q 2 = Q 3, therefore 41.8m + 333m = m = 20930/374.8 = 55.8 g