Respostas

2013-04-29T09:12:33-03:00

Olá Rosilany, temos que:

 

A_o=30\cdot{10}\cdot{5}=1500cm^3

 

A=1800+1800\cdot{0,2\%}=1503cm^3

 

Agora sim, podemos resolver:

 

\Delta{V}=V_o\cdot{3\alpha}\cdot{\Delta{\theta}}

 

Isolando a temperatura:

 

\Delta{\theta}=\frac{\Delta{V}}{V_o\cdot{3\alpha}}

 

\Delta{\theta}=\frac{1503-1500}{1500\cdot{3\cdot{10.10^{-6}}}}

 

\Delta{\theta}=\frac{1503-1500}{1500\cdot{3\cdot{10.10^{-6}}}}

 

Resolvendo:

 

\boxed{\Delta{\theta}=66,67^oC}