First Law of Thermodynamics - Revision history

Laurence12799: /* Examples */

2019-08-04T14:42:29Z

Examples

← Older revision		Revision as of 10:42, 4 August 2019
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	* http://www.shashaiseminar.com/notes/a-level/zimsec/physics/heat-2/		* http://www.shashaiseminar.com/notes/a-level/zimsec/physics/heat-2/
	* https://www.asee.org/public/conferences/8/papers/4611/download		* https://www.asee.org/public/conferences/8/papers/4611/download

	~~==Examples==~~

	~~===Difficult===~~


	~~Solution:~~

	~~<math>∆U_t = Q+W</math>~~


	~~step 1 to 2:~~

	~~<math>∆U_t(12) = -200J, W(12) = -6000J</math>~~

	~~<math>Q(12)=∆U_t(12) - W(12) = -200J - (-6000J) = 5800J</math>~~


	~~step 3 to 4:~~
	~~<math>Q(34) = -800J, W(34) = 300J</math>~~

	~~<math>∆U_t(34)= Q(34) + W(34) = -800J + 300J = -500J</math>~~


	~~<math>∆U_t</math> is a state function.~~
	~~That means a <math>∆U_t</math> for series of steps that leads back to the initial state must be zero.~~
	~~The sum of U for all of the steps must be zero.~~

	~~<math>∆U_t(12341) = 0</math>~~

	~~<math>∆U_t(41) = 4700J</math>~~

	~~<math>∆U_t(23) = 0 - ∆U_t(12) - ∆U_t(34) - ∆U_t(41) = 0 - (-200J) - (-500J) - 4700J = -4000J</math>~~


	~~Step 2 to 3:~~
	~~<math>∆U_t(23) = -4000J, Q(23) = -3800J</math>~~

	~~<math>W(23) = ∆U_t(23) - Q(23) = -4000J - (-3800J) = -200J</math>~~

	~~<math>W(12341) = -1400J</math>~~

	~~<math>W(41) = W(12341) - W(12) - W(23) - W(34) = -1400J - (-6000J) - (-200J) - 300J = 4500J</math>~~


	~~Step 4 to 1:~~
	~~<math>∆U_t(41) = 4700J W(41) = 4500J</math>~~

	~~<math>Q(41) = ∆U_t(41) - W(41) = 4700J - 4500J = 200J</math>~~

Laurence12799: /* Middling */

2019-08-04T14:42:18Z

Middling

← Older revision		Revision as of 10:42, 4 August 2019
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	==Examples==		==Examples==

	~~===Middling===~~

	~~Solution:~~

	~~Utilizing <math>∆KE+∆PE+∆U=Q+W</math>~~

	~~a) No work is done on the system + no work is done by the system to the surroundings, thus <math>W</math> = 0.~~

	~~b) Egg is falling from a higher surface, which means the elevation is decreased and the sign of <math>∆PE</math> is negative (-).~~

	~~c) Because the egg is at rest for both the initial and final state, <math>∆KE</math> = 0.~~

	~~d) The chemical state of egg does not change (it does not get scrambled), thus <math>∆U</math> = 0.~~

	~~e) From the above equation, the values with zero (<math>W</math>, <math>∆PE</math>, and <math>∆KE</math>) can all be cancelled and <math>∆PE</math> would equal <math>Q</math>.~~

	~~<math>∆PE = Q </math>~~

	~~Because the <math>∆PE</math> is negative(-), <math>Q</math> is also negative.~~

	===Difficult===		===Difficult===

Laurence12799: /* Difficult */

2019-08-04T14:41:54Z

Difficult

Show changes

Laurence12799: /* Difficult */

2019-08-03T22:24:15Z

Difficult

← Older revision		Revision as of 18:24, 3 August 2019
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	! style="border-width:3px; border-color:black; background: white" \| <math>\Delta U \ (J)</math>		! style="border-width:3px; border-color:black; background: white" \| <math>\Delta U \ (J)</math>
	! style="border-width:3px; border-color:black; background: white" \| <math>Q \ (J)</math>		! style="border-width:3px; border-color:black; background: white" \| <math>Q \ (J)</math>
	! style="border-width:3px; border-color:black; background: white" \| <math>W \ (J)</math>		! style="border-width:3px; border-color:black; background: white" \| <math>W_{by gas} \ (J)</math>
	\|-		\|-
	\| style="background: white; border-width:3px; border-color:black" \| <math>1 \rightarrow 2</math>		\| style="background: white; border-width:3px; border-color:black" \| <math>1 \rightarrow 2</math>
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	::'''Step''' <math>[1 \rightarrow 2]</math>:		::'''Step''' <math>[1 \rightarrow 2]</math>:

			:::In-between steps 1 and 2, we can see from the table that the internal energy of the gas decreases by 200 Joules, and the gas does 6,000 Joules of Work on the surroundings. Thus, we can solve for the heat added/lost:

			::::<math>Q = \Delta U - W_{on gas} = \Delta U + W_{bygas} = -200 - 6,000 = -6,200 \ J</math>

	::'''Step''' <math>[2 \rightarrow 3]</math>:		::'''Step''' <math>[2 \rightarrow 3]</math>:

Laurence12799: /* Difficult */

2019-08-03T22:11:37Z

Difficult

@@ Line 122: / Line 122: @@
 One mole of gas in a closed system undergoes a four-step Thermodynamic cycle. Using the data below, determine the numerical values for the missing quantities.
-[[File:Thermo3.png|center|400px]]
+{| class="wikitable" style="text-align:center; width:250px; height:250px; font-size:14pt;  border:10; frame:border; rules:all; margin-left:auto; margin-right:auto"
-:'''Step''' <math>[1 \rightarrow 2]</math>:
+The equation we will use for this example is:
-:'''Step''' <math>[2 \rightarrow 3]</math>:
+:<math>\Delta U = Q + W</math>
-:'''Step''' <math>[3 \rightarrow 4]</math>:
+::'''Step''' <math>[1 \rightarrow 2]</math>:
-:'''Step''' <math>[4 \rightarrow 1]</math>:
+::'''Step''' <math>[2 \rightarrow 3]</math>:
-:'''Step''' <math>[1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1]</math>:
+::'''Step''' <math>[3 \rightarrow 4]</math>:
 ==Connectedness==

Laurence12799: /* Difficult */

2019-08-03T21:01:17Z

Difficult

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	===Difficult===		===Difficult===
			One mole of gas in a closed system undergoes a four-step Thermodynamic cycle. Using the data below, determine the numerical values for the missing quantities.

			[[File:Thermo3.png\|center\|400px]]

			:'''Step''' <math>[1 \rightarrow 2]</math>:

			:'''Step''' <math>[2 \rightarrow 3]</math>:

			:'''Step''' <math>[3 \rightarrow 4]</math>:

			:'''Step''' <math>[4 \rightarrow 1]</math>:

			:'''Step''' <math>[1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1]</math>:

	==Connectedness==		==Connectedness==

Laurence12799: /* Difficult */

2019-08-03T20:55:20Z

Difficult

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	===Difficult===		===Difficult===
	~~One mole of gas in a closed system undergoes a four-step thermodynamic cycle. Using the below data, determine numerical values for the missing quantities.~~

	~~[[File:Thermo3.png]]~~

	Solution:		Solution:

Laurence12799: /* Middling */

2019-08-03T20:53:47Z

Middling

← Older revision		Revision as of 16:53, 3 August 2019
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	===Middling===		===Middling===
			An egg, initially at rest, is dropped onto a concrete surface and breaks. Treat the egg as the system.

			:'''a) What is the sign of <math>W</math>?'''

			::Since the egg is speeding up as it falls, Work must be being done on the egg. Thus, <math>W_{onegg}</math> is positive, and <math>W_{byegg}</math> is negative, since:

			:::<math>W_{onegg} = -W_{by egg}</math>

			:'''b) What is the sign of <math>\Delta PE</math>?'''

			::The change in potential energy of the egg will be negative, since it is falling towards the Earth. Let <math>h_0</math> and <math>h_f</math> be the initial and final height of the egg, respectively with <math>h_f << h_0</math>, then we have:

			:::<math>\Delta PE = mgh_f - mgh_0 = mg(h_f - h_0) < 0</math>, since <math>h_f << h_0</math>

			:'''c) What is <math>\Delta KE</math>?'''

			::The change in kinetic energy of the egg is <math>0</math>, due to the fact that its initial and final velocity are both <math>0</math>:

			:::<math>\Delta KE = \frac{1}{2}m{v_f}^2 - \frac{1}{2}m{v_0}^2 = \frac{1}{2}m({v_f}^2 - {v_0}^2) \quad \And \quad v_f = v_0 = 0 \quad \therefore \quad \Delta K = 0</math>

			:'''d) What is <math>\Delta U</math>?'''

			::No Heat is added to the system, and no chemical processes are in motion with the egg. Therefore, we can assume the internal energy of the egg is constant:

			:::<math>\Delta U = 0</math>

			:'''e) What is the sign of <math>Q</math>?'''

			::From our [[First Law of Thermodynamics#Mathematicla Model\| Mathematical Model]], we see:

			:::<math>\Delta PE + \Delta KE + \Delta U = Q + W_{onegg}</math>

			::We also know:

			:::<math>\Delta PE < 0</math>

			:::<math>\Delta KE = 0</math>

			:::<math>\Delta U = 0</math>

			:::<math>W_{onegg} > 0</math>

			::So:

			:::<math>Q = \Delta PE - W_{onegg}</math>

			::We have <math>Q</math> set equal to a negative quantity minus a positive quantity. Therefore, <math>Q</math> must be a negative quantity its self:

			:::<math>Q < 0</math>

	===Difficult===		===Difficult===

Laurence12799: /* Middling */

2019-08-03T20:16:49Z

Middling

← Older revision		Revision as of 16:16, 3 August 2019
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	===Middling===		===Middling===
	~~An egg, initially at rest, is dropped onto a concrete surface and breaks. Treat the egg as the system.~~

	~~a) What is the sign of <math>W</math>?~~

	~~b) What is the sign of <math>∆PE</math>?~~

	~~c) What is <math>∆KE</math>?~~

	~~d) What is <math>∆U</math>?~~

	~~e) What is the sign of <math>Q</math>?~~

	Solution:		Solution:

Laurence12799: /* Simple */

2019-08-03T20:16:05Z

Simple

← Older revision		Revision as of 16:16, 3 August 2019
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	==Examples==		==Examples==

	~~===Simple===~~
	~~Heat of 5000J is added to a closed system and the internal energy is decreased by 11000J. How much energy is transferred as work?~~

	~~Solution:~~

	~~<math>∆U = Q + W</math>, <math>Q= 5000J</math>, <math>∆U=-11000 J</math>~~

	~~<math>-11000J = 5000J + W</math>~~

	~~<math>W = -11000J-5000J</math>~~

	~~<math>W = -16000J</math>~~

	===Middling===		===Middling===