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, Homework Week 10, 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

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See also

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Further reading

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External links

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References

Dean, O. (22-06-2008) Application of Specific Heat Retrieved on 10 November 2015 from http://fiziknota.blogspot.com.au/2008/06/application-of-specific-heat-capacity.html

Breen, M. (2001) Re. What are some practical uses of determining the specific heat of a metal. Retrieved on 10 November 2015 from http://www.madsci.org/posts/archives/2001-02/981315429.Rg.r.html