Coole Cola
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
Need help? Yes, please!
The following quantities appear in the problem:
Masse \(m\) / Temperatur \(T\) / Wärme \(Q\) / spezifische latente Wärme \(L\) / Wärmekapazität \(c\) /
The following formulas must be used to solve the exercise:
\(Q = c \cdot m \cdot \Delta\vartheta \quad \) \(Q = m \cdot L_{\scriptscriptstyle\rm f} \quad \) \(\sum Q^\nearrow \stackrel{!}{=} \sum Q^\swarrow \quad \)
No explanation / solution video to this exercise has yet been created.
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Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Sie wollen ein grosses Glas Cola McO sowie als Wasser zu betrachten! von TcO mit Eiswürfeln CeO von Te abkühlen. Welche Masse an Eiswürfeln müssen Sie dazugeben damit sich eine Mischtemperatur von TmO einstellt? Vernachlässigen Sie jeglichen Wärmeaustausch mit der Umgebung.
Solution:
Geg m_C Mc c_C c_W Cw vartheta_C Tc c_E Ce vartheta_E Te L_fW Lfw vartheta^* Tm GesMasse der Eiswürfelm_Esikg textbfVariante Erster Schritt: Energiebilanz aufstellen Delta Q_Crightarrow + Delta Q_E-rightarrow + Delta Q_f + Delta Q_Erightarrow m_C c_W leftvartheta^*-vartheta_C right+m_E c_E left-vartheta_E right+m_E L_fW+m_E c_W leftvartheta^*- right Diese dann vereinfachen und algebraisch nach gesuchter Grösse auflösen ... m_C c_W leftvartheta^*-vartheta_C right+m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me textbfVariante WaermeSchritte unknownm_ PGleichungDeltasscQC DeltasscQE PGleichungDelta Q_rightarrow Delta Q_-rightarrow + Q_f + Q_rightarrow PGleichungm_C c_W theta^*-theta_C m_E c_E - theta_E + m_E L_f + m_E c_W theta^*- MGleichungm_C c_W theta^*-theta_C -m_E c_E theta_E + m_E L_f + m_E c_W theta^* MGleichungm_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right MGleichungm_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E PHYSMATH m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me
Sie wollen ein grosses Glas Cola McO sowie als Wasser zu betrachten! von TcO mit Eiswürfeln CeO von Te abkühlen. Welche Masse an Eiswürfeln müssen Sie dazugeben damit sich eine Mischtemperatur von TmO einstellt? Vernachlässigen Sie jeglichen Wärmeaustausch mit der Umgebung.
Solution:
Geg m_C Mc c_C c_W Cw vartheta_C Tc c_E Ce vartheta_E Te L_fW Lfw vartheta^* Tm GesMasse der Eiswürfelm_Esikg textbfVariante Erster Schritt: Energiebilanz aufstellen Delta Q_Crightarrow + Delta Q_E-rightarrow + Delta Q_f + Delta Q_Erightarrow m_C c_W leftvartheta^*-vartheta_C right+m_E c_E left-vartheta_E right+m_E L_fW+m_E c_W leftvartheta^*- right Diese dann vereinfachen und algebraisch nach gesuchter Grösse auflösen ... m_C c_W leftvartheta^*-vartheta_C right+m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me textbfVariante WaermeSchritte unknownm_ PGleichungDeltasscQC DeltasscQE PGleichungDelta Q_rightarrow Delta Q_-rightarrow + Q_f + Q_rightarrow PGleichungm_C c_W theta^*-theta_C m_E c_E - theta_E + m_E L_f + m_E c_W theta^*- MGleichungm_C c_W theta^*-theta_C -m_E c_E theta_E + m_E L_f + m_E c_W theta^* MGleichungm_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right MGleichungm_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E PHYSMATH m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me
Meta Information
Exercise:
Sie wollen ein grosses Glas Cola McO sowie als Wasser zu betrachten! von TcO mit Eiswürfeln CeO von Te abkühlen. Welche Masse an Eiswürfeln müssen Sie dazugeben damit sich eine Mischtemperatur von TmO einstellt? Vernachlässigen Sie jeglichen Wärmeaustausch mit der Umgebung.
Solution:
Geg m_C Mc c_C c_W Cw vartheta_C Tc c_E Ce vartheta_E Te L_fW Lfw vartheta^* Tm GesMasse der Eiswürfelm_Esikg textbfVariante Erster Schritt: Energiebilanz aufstellen Delta Q_Crightarrow + Delta Q_E-rightarrow + Delta Q_f + Delta Q_Erightarrow m_C c_W leftvartheta^*-vartheta_C right+m_E c_E left-vartheta_E right+m_E L_fW+m_E c_W leftvartheta^*- right Diese dann vereinfachen und algebraisch nach gesuchter Grösse auflösen ... m_C c_W leftvartheta^*-vartheta_C right+m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me textbfVariante WaermeSchritte unknownm_ PGleichungDeltasscQC DeltasscQE PGleichungDelta Q_rightarrow Delta Q_-rightarrow + Q_f + Q_rightarrow PGleichungm_C c_W theta^*-theta_C m_E c_E - theta_E + m_E L_f + m_E c_W theta^*- MGleichungm_C c_W theta^*-theta_C -m_E c_E theta_E + m_E L_f + m_E c_W theta^* MGleichungm_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right MGleichungm_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E PHYSMATH m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me
Sie wollen ein grosses Glas Cola McO sowie als Wasser zu betrachten! von TcO mit Eiswürfeln CeO von Te abkühlen. Welche Masse an Eiswürfeln müssen Sie dazugeben damit sich eine Mischtemperatur von TmO einstellt? Vernachlässigen Sie jeglichen Wärmeaustausch mit der Umgebung.
Solution:
Geg m_C Mc c_C c_W Cw vartheta_C Tc c_E Ce vartheta_E Te L_fW Lfw vartheta^* Tm GesMasse der Eiswürfelm_Esikg textbfVariante Erster Schritt: Energiebilanz aufstellen Delta Q_Crightarrow + Delta Q_E-rightarrow + Delta Q_f + Delta Q_Erightarrow m_C c_W leftvartheta^*-vartheta_C right+m_E c_E left-vartheta_E right+m_E L_fW+m_E c_W leftvartheta^*- right Diese dann vereinfachen und algebraisch nach gesuchter Grösse auflösen ... m_C c_W leftvartheta^*-vartheta_C right+m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right m_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me textbfVariante WaermeSchritte unknownm_ PGleichungDeltasscQC DeltasscQE PGleichungDelta Q_rightarrow Delta Q_-rightarrow + Q_f + Q_rightarrow PGleichungm_C c_W theta^*-theta_C m_E c_E - theta_E + m_E L_f + m_E c_W theta^*- MGleichungm_C c_W theta^*-theta_C -m_E c_E theta_E + m_E L_f + m_E c_W theta^* MGleichungm_Eleft-c_E vartheta_E+ L_fW+ c_W vartheta^*right -m_C c_W leftvartheta^*-vartheta_C right MGleichungm_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E PHYSMATH m_E frac-m_C c_W leftvartheta^*-vartheta_C rightL_fW+c_W vartheta^*-c_E vartheta_E Me
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Asked Quantity:
Masse \(m\)
in
Kilogramm \(\rm kg\)
Physical Quantity
Eigenschaft der Materie
Unit
Base?
SI?
Metric?
Coherent?
Imperial?