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masha68 [24]
3 years ago
12

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

istor is 3,650 °C. Find: If the thermistor is then used to measure a particular temperature, and its resistance is measured as 500 Ω, determine the thermistor temperature.
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

the thermistor temperature = 325.68 \ ^0 \ C

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

Resistance of the thermistor R_1 = 20,000 ohms

Material constant \beta = 3650

Resistance of the thermistor R_2 = 500 ohms

Using the equation :

R_1 = R_2  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{R_1}{ R_2} =   \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

Taking log of both sides

In \ \frac{R_1}{ R_2} = In \  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

\frac{1}{T_2} =   \frac{1}{T_1}  -          \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

{T_2} =  \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

Replacing our values into the above equation :

{T_2} =  \frac{3650*373}{3650 - In (\frac{20000}{500})373}

{T_2} =  \frac{1361450}{3650 - 3.6888*373}

{T_2} =  \frac{1361450}{3650 - 1375.92}

{T_2} =  \frac{1361450}{2274.08}

{T_2} = 598.68 \ K

{T_2} = 325.68 \ ^0 \ C

Thus, the thermistor temperature = 325.68 \ ^0 \ C

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a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

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Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

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c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

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