Now that we are familiar with heat transfer methods, we need to examine what occurs when objects either absorb or lose heat.

 

According to the Zeroth Law of Thermodynamics, heat flows from an object at a high temperature to one at a lower temperature until they reach thermal equilibrium - the same temperature.

 

This naturally leads us to the Law of Heat Exchange,

 

ΔQ_lost+ ΔQ_gained = 0

 

where, in the absence of a phase change, ΔQ = mcΔT. In this formula,

 

ΔQ is the heat lost or gained

m is the mass of the object in kilograms

ΔT is the object's temperature change (T_f - T_o) in either Celsius or Kelvin

c is the object's specific heat in J/kgK

 

For example, water has a specific heat of 4186 J/kgK. This tells us that a kilogram (1000 ml) of water requires 4186 J of heat to increase the temperature of the sample 1 ºC or 1 Kelvin degree.

 

Table courtesy of http://faculty.wwu.edu/~vawter/physicsnet/topics/Thermal/HeatCapTable.html

 

To show how the Law of Heat Exchange is utilized in problems, let's say that we combine 50 ml of water at 20ºC  with 50 ml of water at 100ºC. What will be the equilibrium temperature of the mixture?

 

ΔQ_lost+ ΔQ_gained = 0

 

0.050(4186)(T_f - 100) + 0.050(4186)(T_f - 20) = 0

209.3(T_f - 100) + 209.3(T_f - 20) = 0

209.3T_f - 20930 + 209.3T_f - 4186 = 0

418.6T_f - 25116 = 0 

418.6T_f = 25116

T_f = 60ºC

 

This result seems obvious since there were equal quantities of water mixed together - it is merely the average of the two original temperatures. But what if there was a difference in the quantities being mixed or different materials were involved?

 

Suppose a 300 gram sample of aluminum is heated to 100ºC by placing it in a beaker of boiling water. Later it is completely submerged in a Styrofoam cup containing 100 grams of water at 20ºC. What will be the equilibrium temperature?

 

ΔQ_lost+ ΔQ_gained = 0

 

0.300(900)(T_f - 100) + 0.100(4186)(T_f - 20) = 0

270(T_f - 100) + 418.6(T_f - 20) = 0

270T_f - 27000 + 418.6T_f - 8372 = 0

688.6T_f  - 35372 = 0

688.6T_f  = 35372

T_f  = 51.4ºC

 

This result is not easily guessed but can be verified experimentally if the correct materials are available.

 

Sometimes when heat is lost or gained the sample changes phase or state. When this occurs, we use two other formulas:

 

Q = mL_f     or     Q = mL_v

 

where L_f and L_v are called the latent heat of fusion and the latent heat of vaporization. Notice the absence of any temperature value in either formula. This is because the term "latent" refers to "hidden" heat. The heat is "hidden" because there is no observable temperature change. Latent heats affect the potential energy of the molecules in the sample (when they are transitioning between their solid and liquid states at their melting point, or between their liquid and vapor states at their the boiling point) whereas, temperature changes reflect a change in the kinetic energy of the sample's molecules. For water these values are: L_f = 334 kJ/kg and L_v = 2300 kJ/kg.

 

Image courtesy of Hecht Physics (Algebra/Trig) 1994

 

In the image shown above, heat is being added at a steady rate to a 1 kilogram sample of ice at -10ºC. The flat lines are where the water molecules are changing state; that is, going through a phase change. The relative lengths of these plateaus show how the magnitudes of these hidden heats compare - the plateau for vaporization is roughly 7 times longer than the one for fusion. The slanted lines represent changes in the sample's temperature. During these sections the ice is warming, the water is warming, and the steam is superheating.

 

As am example, suppose that 500 grams of ice at -10ºC is heated until it becomes steam at 110º. Heat is being added to the system at a rate of 600 J/sec.

 

transition	formula	heat needed	seconds	minutes
warming ice	ΔQ = mcΔT	Q = 0.5(2093)(10)	17.4	0.290
melting ice	Q = mLf	Q = 0.5(334,000)	278.3	4.64
warming water	ΔQ = mcΔT	Q = 0.5(4186)(100)	348.8	5.81
vaporizing water	Q = mLv	Q = 0.5(2,300,000)	1916.7	31.9
heating steam	ΔQ = mcΔT	Q = 0.5(2009)(10)	16.7	0.279

 

 

In the above graph, steeper slopes represent smaller specific heats. This should make sense since a smaller specific heat means that the substance requires less heat to make a temperature change. Ice and steam have specific heats that are very close to the same magnitude. Water's specific heat is greater than either by at least a factor of 2.

 

Later when we study gases and thermodynamics processes, we will learn about molar specific heats for constant pressure and constant volume, C_P and C_V.