Energy Transfer due to a Temperature Difference: Difference between revisions

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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.


==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'''<br>
<b>delta(E) = Q + W</b>
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:
<b>The Energy Principle</b>
delta(E_system) = W + Q + other energy transfers
<b>Q</b>
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. <br>
[https://www.youtube.com/embed/bODiX2PjCPE Heat Transfer]
Check this video out for a detailed look into the topic.<br>
[https://www.youtube.com/embed/XmXr58ySUhI Relationship between deltaE and Q+W]
==Examples==
<b>Example 1:</b>
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
<b>Example 2:</b>
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.
<b>Example 3:</b>
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
<h>
Example 3:
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Latest revision as of 13:07, 6 August 2019