diff --git a/src/1_einstein_model_solutions.md b/src/1_einstein_model_solutions.md
index 4e9436a8dfd9c07f1734361c2bd104624b15d90d..e39e817df332b53ead70c0ba37e4b3ce61003c22 100644
--- a/src/1_einstein_model_solutions.md
+++ b/src/1_einstein_model_solutions.md
@@ -1,66 +1,71 @@
 # Solutions for lecture 1 exercises
+### Warm-up exercises
+1. An ideal gas only contains 3 positional degrees of freedom.
+2. $C = 2k_B$.
+3. See image below
+```python
+import matplotlib.pyplot as plt
+import numpy as np
+
+fig, ax = plt.subplots()
+omega = np.linspace(0.1, 3)
+T = [1,2]
+ax.plot(omega, 1/(np.exp(omega/T[0]) - 1), 'b-', label = r'$T_1$')
+ax.plot(omega, 1/(np.exp(omega/T[1]) - 1), 'r-', label = r'$T_2$')
+ax.set_ylim([0,3])
+ax.set_xlim([0,3])
+ax.set_xlabel('$\hbar \omega$')
+ax.set_xticks([0])
+ax.set_xticklabels(['$0$'])
+ax.set_ylabel('$n_B$')
+ax.set_yticks([0,1, 2])
+ax.set_yticklabels(['$0$','$1$', '$2$'])
+ax.set_title('$T_1<T_2$')
+ax.legend()
+draw_classic_axes(ax, xlabeloffset=.2);
+```
+4. Minus sign in the exponent. This would result in $n_B(T = 0) = -1$, which is not physical.
+5. See plot with slider
 
 ### Exercise 1: Heat capacity of a classical oscillator.
 
-1.
-
-$$
+1. $$
 Z = \int_{-\infty}^{\infty}dp \int_{-\infty}^{\infty} dx e^{-\frac{\beta}{2m}p^2-\frac{\beta k}{2}x^2} = \sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta k}} = \frac{2\pi}{\beta}\sqrt{\frac{m}{k}},
 $$
 
 where we used $\int_{-\infty}^{\infty}e^{-\alpha x^2} = \sqrt{\frac{\pi}{\alpha}}$.
 
-2.
-
-$$
+2. $$
 \langle E \rangle = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = \frac{1}{\beta}
 $$
 
-3.
-
-$$
+3. $$
 C = \frac{\partial\langle E\rangle}{\partial T} = k_B
 $$
-
-4.
-
-Since this is a 1D system, the prefactor is $1$ as expected.
+The heat capacity is temperature independent.
 
 ### Exercise 2: Quantum harmonic oscillator.
 
-1.
-
-Take a look into the graph from the notes.
-
-2.
-
-$$
+1. $$
 Z = \sum_{n = 0}^{\infty} e^{-\beta\hbar\omega(n + 1/2)} = e^{-\beta\hbar\omega/2}\frac{1}{1 - e^{-\beta\hbar\omega}} = \frac{1}{2\sinh(\beta\hbar\omega/2)},
 $$
 
 where we used $\sum_{n = 0}^{\infty}r^n = \frac{1}{1 - r}$.
 
-3.
-
-$$
+2. $$
 \langle E\rangle = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = \frac{\hbar\omega}{2}\coth\frac{\beta\hbar\omega}{2} = \hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right) = \hbar\omega\left(n_B(\beta\hbar\omega) + \frac{1}{2}\right).
 $$
 
-4.
-
-$$
+3. $$
 C = \frac{\partial \langle E\rangle}{\partial T} = \frac{\partial\langle E\rangle}{\partial\beta}\frac{\partial\beta}{\partial T} = k_B(\beta\hbar\omega)^2\frac{e^{\beta\hbar\omega}}{(e^{\beta\hbar\omega} - 1)^2}.
 $$
 
 In the high temperature limit $\beta \rightarrow 0$ and $e^{\beta\hbar\omega} \approx 1 + \beta\hbar\omega$, so $C \rightarrow k_B$ which is the same result as in Exercise 1.3.
 
-5.
-
-We can see that for $\beta\hbar\omega = 1$ we already have $C \approx 0.92k_B$, so we can define a characteristic temperature $T_E = \hbar\omega/k_B$ such that temperatures above this one can be considered high.
-
-6.
+4. Compare your result with the plot with the slider. 
+Did you correctly indicate the where the Einstein temperature is?
 
-$$
+5. $$
 \langle n\rangle = \frac{1}{Z}\sum_{n = 0}^{\infty} ne^{-\beta\hbar\omega(n + 1/2)} = 2\frac{e^{\beta\hbar\omega/2} - e^{-\beta\hbar\omega/2}}{2}e^{-\beta\hbar\omega/2}\frac{e^{-\beta\hbar\omega}}{(1 - e^{-\beta\hbar\omega})^2} = \frac{1}{e^{\beta\hbar\omega} - 1},
 $$
 
@@ -68,29 +73,21 @@ where we used $\sum_{n = 0}^{\infty}nr^n = \frac{r}{(1 - r)^2}$.
 
 ### Exercise 3: Total heat capacity of a diatomic material.
 
-1.
-
-Use the formula $\omega = \sqrt{\frac{k}{m}}$.
-
-2.
+1. Use the formula $\omega = \sqrt{\frac{k}{m}}$.
 
-Energy per atom is given by 
+2. Energy per atom is given by 
 
 $$
 E = \frac{N_{^6Li}}{N}\hbar\omega_{^6Li}(2 + 1/2) + \frac{N_{^7Li}}{N}\hbar\omega_{^7Li}(4 + 1/2).
 $$
 
-3.
-
-Energy per atom is given by
+3. Energy per atom is given by
 
 $$
 E = \frac{N_{^6Li}}{N}\hbar\omega_{^6Li}\left(n_B(\beta\hbar\omega_{^6Li}) + \frac{1}{2}\right) + \frac{N_{^7Li}}{N}\hbar\omega_{^7Li}\left(n_B(\beta\hbar\omega_{^7Li}) + \frac{1}{2}\right).
 $$
 
-4.
-
-Heat capacity per atom is given by
+4. Heat capacity per atom is given by
 
 $$
 C = \frac{N_{^6Li}}{N}C_{^6Li} + \frac{N_{^7Li}}{N}C_{^7Li},