Respuesta :
Answer:
[tex]68.3^{\circ}C[/tex]
Explanation:
The relationship between voltage, resistance and current in a metal wire is given by
[tex]V=RI[/tex]
where
V is the voltage
R is the resistance
I is the current
Moreover, the resistance of a metal wire as a function of the temperature is:
[tex]R(T)=R_0(1+\alpha(T-T_0))[/tex] (1)
where
[tex]R_0[/tex] is the resistance at temperature [tex]T_0[/tex]
[tex]\alpha[/tex] is the temperature coefficient
In this problem, at 30.0∘C, we have
V = 10.00 V
I = 0.4076 A
So the resistance at this temperature is
[tex]R_0=\frac{V_0}{I_0}=\frac{10.00}{0.4076}=24.53 \Omega[/tex]
Then, at the unknown temperature T, the current is
I = 0.3502 A
So the resistance is
[tex]R(T)=\frac{V}{I}=\frac{10.00}{0.3502}=28.56\Omega[/tex]
Substituting and solving for (1), we find the unknown temperature:
[tex]T=T_0+\frac{\frac{R}{R_0}-1}{\alpha}=30.0+\frac{\frac{28.56}{24.53}-1}{0.00429}=68.3^{\circ}C[/tex]
The value of temperature in new environment is 68 degree Celsius.
The resistance is defined as ratio of voltage to the current.
Resistance, [tex]R=\frac{V}{I}[/tex]
When voltage is 10.00-V power supply, and a current of 0.4076 A .
Resistance, [tex]R_{1}=\frac{10}{0.4076}=24.54ohm[/tex]
When voltage is 10.00-V power supply, and a current of 0.3502A
[tex]R_{2}=\frac{10}{0.3502}=28.55[/tex] ohm
Here we use relationship,
[tex]R_{2}=R_{1}(1+\alpha (T_{2}-T_{1}))[/tex] and [tex]T_{1}=30[/tex]
Where [tex]\alpha[/tex] is the temperature coefficient of resistivity.
Substituting all values in above equation
[tex]28.55=24.54(1+0.00429(T_{2}-30))\\\\1+0.00429(T_{2}-30)=1.1634\\\\0.00429(T_{2}-30)=1.1634-1=0.1634\\\\(T_{2}-30)=38.09\\\\T_{2}=38.09+30=68.09[/tex]
Hence, The value of temperature in new environment is 68 degree Celsius.
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