Energy Transfer due to a Temperature Difference

This section will be sharing how to calculate energy transfer in systems that are affected by temperature changes. This related to Chapter 7 in the Text Book and Exam 3.

Woong Jun Park
wpark39

The Main Idea

When hot and cold objects are placed into contact with one another, there is a transfer of energy from the hot to the cold object. This is not your typical energy transfer as "work", but rather it is called Q.

Q = Energy Transfer Due to a Temperature Difference
delta(E) = Q + W

Like Work (W), Q can be negative because there could be a transfer of energy out of the system rather than it coming into the system (this can happen if the system has a higher temperature than its surroundings).

In the following, we will cover examples and more functionalities of how this Energy Transfer works.

A Mathematical Model

The following are mathematical models that you will use in these calculations:

The Energy Principle delta(E_system) = W + Q + other energy transfers

Q Q = mC(deltaT) Q = Heat Added m = mass C = Specific Heat deltaT = change in Temperature

A Computational Model

Check this video out for a brief understanding of how Heat & Temperature relate to Physics.
Heat Transfer

Check this video out for a detailed look into the topic.
Relationship between deltaE and Q+W

Examples

Example 1: How much heat energy is required to raise the temperature of 55.0g of water from 25° C to 28.6° C?

Q=mCdeltatT
Q=55g*4.2J/g/K*(28.6° C-25° C)
Q=827.64J

Example 2: 180 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 800 grams and initial temperature 24° C (the heat capacity of aluminum is 0.9 J/gram/K).

After a short time, what is the temperature of the water?

System: Water+Pan
Q = mC(deltaT)
180g*4.2(J/g/K)*(100-T) = 800g*0.9(J/g/K)*(T-22)
T=61.95° C

• In this case, you would assume that the heat capacities for both water and aluminum don't really change with temperature. And also assume the energy transfer between the system and the surroundings was negligible.
  

Example 3: Suppose you warm up 510 grams of water (about half a liter, or about a pint) on a stove, and while this is happening, you also stir the water with a beater, doing 6104 J of work on the water. After the large-scale motion of the water has dissipated away, the temperature of the water is observed to have risen from 23°C to 82°C.
A) What was the change in the thermal energy of the water?
deltatE = mCdeltaT
510g*4.2(J/g/K)(82-23)
deltatE = 126378 J

B) Taking the water as the system, how much energy transfer due to a temperature difference (microscopic work) Q was there across the system boundary?
Q = E-W
126378J - 6e4J
Q = 66378J

C) Taking the water as the system, what was the energy change of the surroundings?
deltaE_system = - deltaE_surroundings
deltaE_surroundings = -126378

Connectedness

I am a Business, Pre-Dental student so health topics have always been very interesting to me. Although this topic may not relate directly to my major, it does have real world applications in the health field.

Specific heat and this transfer of energy can be applied to taking patient's temperatures using a thermometer. There is liquid and material that is inside the thermometer that a have low specific capacities. This allows for a very precise measurement of the temperature. This is not necessarily

Another industrial application can e seen in engine parts that expand and contract because of the constant change in heat within the engine.

History

The history of heat and work can be dated back to 1789 when the French scientist Antoine Lavoisier created a new theory - the phlogiston theory on Chemistry that basically negated all previous findings of combustion. He gave heat a meaning and definition that led to the interpretation of heat that is accepted today.

The modern interpretation is what we hold true in the example questions that we used earlier Q = mCdeltatT.

In 1798, Benjamin Thompson - minister for war and police in the German state of Bavaria - wanted to figure out where all the heat from the cannons were coming from. He observed that the surroundings of the cannon got hotter and not colder. He hypothesized that some of the mechanical work done on the cannon was converted to heat.

In 1849, English physicist James Prescott Joule published his work and findings on the conversion of work to heat that Thompson started. He formulated work equivalent of heat. 1 newton meter of work = 0.241 calories of heat.

In 1850, German physicist Clausius published his works on how conserved quantity is neither heat nor work, but a combination of both. He named this Energy and that is where we get the macroscopic equation deltaE = Q - W.

See also

Relevant material and additional reading can be looked into the history of Joule and the laws of thermodynamics.

Further reading

Specific Heat: Britannica
Dependence of Heat Capacity on Thermal History: Nature Publishing Group
Environmental Research Web | Uneven Climate Change Due to Atmospheric Heat Capacity

External links

USGS Water Science School
Science Direct | Determination of specific heat capacity by transient plane source

References

"A Brief History of Heat and Work." A Brief History of Heat and Work. N.p., n.d. Web. 10 Nov. 2015.\

"Application of Specific Heat Capacity" : Application of Specific Heat Capacity. N.p., n.d. Web. 10 Nov. 2015.

"CHAPTER 4: HEAT." : 4.2 Specific Heat Capacity. N.p., n.d. Web. 10 Nov. 2015.

Division of Building Technology, Chalmers University of Technology, Gothenburg SE-412 96, Sweden Received 23 August 2013, Revised 3 September 2013, Accepted 10 September 2013, Available online 22 October 2013

"Re: What Are Some Practical Uses of Determining the Specific Heat of a Metal?" Re: What Are Some Practical Uses of Determining the Specific Heat of a Metal? N.p., n.d. Web. 10 Nov. 2015.