# Heat and Internal Energy

## Presentation on theme: "Heat and Internal Energy"— Presentation transcript:

Heat and Internal Energy
Consider two isolated objects which are insulated from their environments (no heat flow in/out) with temperatures T1 < T2 If they are brought into contact (still insulated from their environment), heat will flow from the warmer object to the cooler one until their temperatures are equal, T1 < Tf < T2 Q=Heat , units of J Heat is the transfer of internal energy from T1 T2 Q T2 T1

the warmer object to the cooler object
The transfer of heat depends on the properties of the objects (T, m, and one other) The heat that is gained or loss is given by c is the Specific Heat, it depends on the substance (see table 16-2), has units of J/(kg C°) Assuming that there are no loses/gains of heat to or from the environment (a statement of conservation of energy), all the heat loss by object 2 must equal the heat gained by object 1

Let’s let the objects come to equilibrium
Solve for Tf

Now, let us assume the objects are made of the same material and have the same mass
Just the average temperature Or lets say object 2 is a cup of hot coffee, while object 1 is a roomful of cool air So coffee cools to room temperature. Returning to the Specific Heat, solids, liquids, and gases have specific heats c, but in

But gases, can be compressed
general c depends upon whether the process (change in temperature) occurs under conditions of constant pressure or constant volume. However, as solids and liquids are generally incompressible, we can ignore this difference. But gases, can be compressed cP – specific heat under constant pressure cV – specific heat under constant volume In fact, the ratio of cP to cV is the adiabatic index of a gas Given here for N2

Latent Heat (17.5-6) The addition or removal of heat from a substance can also result in a change of phase Gas condense evaporate Liquid condense sublimate freeze melt Solid However, phase changes occur at a constant temperature

For example, by adding heat to a pot of water, we know it boils (evaporates) at 100 C. This means that the temperature of both the liquid water and the water vapor (gas) is 100 C. We can add more heat, but the temperature remains constant until all the liquid is evaporated. If we collected the steam, we could add heat to it, then its temperature could increase above 100 C. The heat (or energy) required to make a phase change is called the Latent Heat Latent Heat = L, units of [J/kg] Latent heats for various substances are given in Table 17-4

Example Problem Solid  Liquid Latent Heat of Fusion Lf
Liquid  Gas Latent Heat of Vaporization Lv Solid  Gas Latent Heat of Sublimation Ls The actual heat addition or removal required to cause a phase change Example Problem Water at 23 °C is sprayed on kg of molten gold at 1063 °C (its melting point). The water boils away leaving solid gold at 1063 °C. What minimum mass of water must be used?

Solution: Given: Tow=23°C, Tog=1063°C=Tfg (=melting temp.), Tfw=100°C, mg=0.180 kg Approach: 1) water is heated from 23°C to 100°C – we need specific heat of water, 2) at 100°C the water boils – we need latent heat of vaporization of water, 3) the heat which is added to the water is the heat removed from the gold as it makes a phase transition from liquid to solid – we need the latent heat of fusion for gold, and 4) if we assume no heat losses, energy must be conserved. From Tables 16-2 and 17-4

Heat to raise temperature of water
Heat to evaporate water Heat removed to solidify gold By energy conservation

But how is the heat transferred from the gold to the water
But how is the heat transferred from the gold to the water? It does not happened instantaneously. What about the area of contact?