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Answer:
The final temperature of the element = 262.67°C
The power dissipated in the heating element initially = 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W
Explanation:
Our given parameters include;
A Nichrome heating element operates on 120 V.
Voltage (V) = 120V
Initial Current (I₁) = 1.36 A
Initial Temperature (T₁) = 28°C
Final Current (I₂) = 1.23 A
Final Temperature (T₂) = unknown ????
Temperature dependencies of resistance is given by:
---------------------- (1)
in which R₁ is the resistance at temperature T₁
is the resistance at temperature T₂
Given that V= IR
R =
Therefore, the resistance at temperature 28°C is;
= 88.24Ω
=
= 97.56Ω
From (1) above;
97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]
1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)
0.1056 = 4.5×10⁻⁴(T₂-28°C)
T - 28° C = 234.67
T = 234.67 + 28° C
T = 262.67 ° C
(b)
What is the power dissipated in the heating element initially and when the current reaches 1.23 A
The power dissipated in the heating element initially can be calculated as:
P = I²₁R₂₈
P = (1.36A)²(88.24Ω)
P = 163.209 W
P ≅ 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:
P = (1.23)²(97.56Ω)
P = 147.598524
P ≅ 147.60 W
The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:
where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get
Therefore. the elastic potential energy stored in the spring is 6.75J .
Learn more about potential energy here:
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