Answer
See the detailed answer below.
Work Step by Step
The given figure shows 5 stages of motion:
1) Heating the solid phase sample until it reaches the temperature of melting $Q_1=\bf 20\;\rm kJ$ (the first tilting straight line).
2) Melting the whole sample at constant temperature $Q_2=40-20=\bf 20\;\rm kJ$ (the first horizontal line) until it converts to a liquid phase.
3) Heating the liquid phase sample to the boiling point (the second tilting straight line) $Q_3=120-40=\bf 80\;\rm kJ$.
4) Evaporating the whole sample at constant temperature $Q_4=180-120=\bf 60\;\rm kJ$ (the second horizontal line) until it converts to a vapor phase.
5) Heating the vapor phase sample until it reaches the temperature of 60$^\circ$C $Q_5=200-180=\bf 20\;\rm kJ$ (the third tilting straight line).
Now we can easily answer the questions:
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a) We need to use the first stage of heating:
$$Q_1=mc_{solid}\Delta T_1$$
Thus,
$$c_{solid}=\dfrac{Q_1}{m\Delta T_1}$$
Plugging the known;
$$c_{solid}=\dfrac{20\times 10^3}{(0.5)[-20-(-40)]}$$
$$c_{solid}=\color{red}{\bf 2000}\;\rm J/kg\cdot K$$
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b) By the same approach, we need to use the third stage:
$$c_{liquid}=\dfrac{Q_3}{m\Delta T_3}$$
Plugging the known;
$$c_{liquid}=\dfrac{80\times 10^3}{(0.5)[40-(-20)]}$$
$$c_{liquid}=\color{red}{\bf 2667}\;\rm J/kg\cdot K$$
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c) Melting temperature is at stage 2, the first horizontal line which is
$$T_{melting}=\color{red}{\bf -20}^\circ\;\rm C$$
And the boiling temperature is at stage 4, the second horizontal line which is
$$T_{boiling }=\color{red}{\bf 40}^\circ\;\rm C$$
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d) Latent heat of fusion is given from the second stage:
$$L_{\rm f}=\dfrac{Q_2}{m}$$
Plugging the known;
$$L_{\rm f}=\dfrac{20\times 10^3}{0.5}$$
$$L_{\rm f}=\color{red}{\bf 40\times 10^3}\;\rm J/kg $$
Latent heat of vaporization is given from the fourth stage:
$$L_{\rm v}=\dfrac{Q_4}{m}$$
Plugging the known;
$$L_{\rm v}=\dfrac{60\times 10^3}{0.5}$$
$$L_{\rm v}=\color{red}{\bf 120\times 10^3}\;\rm J/kg $$
