Đáp án:

 13:D

14:B

15:B

16:D

17:C

18: $Q=136,4J$

Giải thích các bước giải:

 Câu 13: ${{m}_{Al}}=0,21kg;{{t}_{Al}}={{200}^{0}}C;{{t}_{nc}}={{30}^{0}}C;{{t}_{cb}}={{50}^{0}}C$

khi có sự cân bằng nhiệt:

$\begin{align}
  & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & \Leftrightarrow {{m}_{Al}}.{{c}_{Al}}.({{t}_{Al}}-{{t}_{cb}})={{m}_{nc}}.{{c}_{nc}}.({{t}_{cb}}-{{t}_{nc}}) \\ 
 & \Leftrightarrow 0,21.880.(200-50)={{m}_{nc}}.4200.(50-30) \\ 
 & \Rightarrow {{m}_{nc}}=0,33kg \\ 
\end{align}$

Câu 14: 

${{m}_{dong}}=0,6kg;{{t}_{d}}={{120}^{0}}C;{{m}_{nc}}=0,5kg;{{t}_{nc}}={{20}^{0}}C$

Khi có sự cân bằng nhiệt:

$\begin{align}
  & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & \Leftrightarrow {{m}_{d}}.{{c}_{d}}.({{t}_{d}}-{{t}_{cb}})={{m}_{nc}}.{{c}_{nc}}.({{t}_{cb}}-{{t}_{nc}}) \\ 
 & \Leftrightarrow 0,6.400.(120-{{t}_{cb}})=0,5.4200.({{t}_{cb}}-20) \\ 
 & \Rightarrow {{t}_{cb}}=30,{{26}^{0}}C \\ 
\end{align}$

Câu 15:

${{m}_{1}}=0,2kg;{{t}_{1}};{{m}_{2}}=0,28kg;{{t}_{2}}={{20}^{0}}C;{{t}_{3}}={{80}^{0}}C$

Khi cân bằng nhiệt xảy ra: 

$\begin{align}
  & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & \Leftrightarrow {{m}_{1}}.{{c}_{1}}.({{t}_{1}}-{{t}_{3}})={{m}_{2}}.{{c}_{2}}.({{t}_{3}}-{{t}_{2}}) \\ 
 & \Leftrightarrow 0,2.400.({{t}_{1}}-80)=0,28.4200.(80-20) \\ 
 & \Rightarrow {{t}_{1}}={{962}^{0}}C \\ 
\end{align}$

Câu 16: 

${{m}_{Al}}=0,5kg;{{m}_{nc}}=0,118kg;{{t}_{0}}={{20}^{0}}C;{{m}_{Fe}}=0,2kg;{{t}_{Fe}}={{75}^{0}}C$

Khi có sự cân bằng xảy ra: 

$\begin{align}
  & {{m}_{Al}}=0,5kg;{{m}_{nc}}=0,118kg;{{t}_{0}}={{20}^{0}}C;{{m}_{Fe}}=0,2kg;{{t}_{Fe}}={{75}^{0}}C \\ 
 & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & \Leftrightarrow {{m}_{Fe}}.{{c}_{Fe}}.({{t}_{Fe}}-{{t}_{cb}})=\left( {{m}_{Al}}.{{c}_{Al}}+{{m}_{nc}}.{{c}_{nc}} \right).({{t}_{cb}}-{{t}_{0}}) \\ 
 & \Leftrightarrow 0,2.460.(75-{{t}_{cb}})=(0,5.896+0,118.4180).({{t}_{cb}}-20) \\ 
 & \Rightarrow {{t}_{cb}}={{25}^{0}}C \\ 
\end{align}$

Câu 17: 

$p=100N/{{m}^{2}};{{V}_{1}}=4{{m}^{3}};{{t}_{1}}={{57}^{0}}C;{{t}_{2}}={{87}^{0}}C;$

đẳng áp:

$\begin{align}
  & \dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}} \\ 
 & \Leftrightarrow \dfrac{4}{57+273}=\dfrac{{{V}_{2}}}{87+273} \\ 
 & \Rightarrow {{V}_{2}}=4,364{{m}^{3}} \\ 
\end{align}$

công do khối khí thực hiện:

$A=p.({{V}_{2}}-{{V}_{1}})=100.(4,364-4)=36,4J$

Câu 18:

$p=100N/{{m}^{2}};{{V}_{1}}=4{{m}^{3}};{{t}_{1}}={{57}^{0}}C;{{t}_{2}}={{87}^{0}}C;$

đẳng áp:

$\begin{align}
  & \frac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}} \\ 
 & \Leftrightarrow \dfrac{4}{57+273}=\dfrac{{{V}_{2}}}{87+273} \\ 
 & \Rightarrow {{V}_{2}}=4,364{{m}^{3}} \\ 
\end{align}$

công do khối khí thực hiện:

$A=p.({{V}_{2}}-{{V}_{1}})=100.(4,364-4)=36,4J$

Nội năng của khối khí:

$p=100N/{{m}^{2}};{{V}_{1}}=4{{m}^{3}};{{t}_{1}}={{57}^{0}}C;{{t}_{2}}={{87}^{0}}C;$

đẳng áp:

$\begin{align}
  & \dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}} \\ 
 & \Leftrightarrow \dfrac{4}{57+273}=\dfrac{{{V}_{2}}}{87+273} \\ 
 & \Rightarrow {{V}_{2}}=4,364{{m}^{3}} \\ 
\end{align}$

công do khối khí thực hiện:

$A=p.({{V}_{2}}-{{V}_{1}})=100.(4,364-4)=36,4J$