If a certain silver wire has a resistance of 8.60 Ω at 29.0°C, what resistance will it have at 42.0°C?

9.027 Ω

Explanation:

Using,

R = R₀(1+αΔt)

R = R₀(1+α[t₂-t₁]) Equation 1

Where R = the value of the resistance at the final temperature, R₀ = the value of the resistance at the initial temperature, α = Temperature coefficient of resistance, t₂ = Final temperature, t₁ = Initial temperature.

Given: R₀ = 8.6 Ω, t₂ = 42 °C, t₁ = 29 °C

Constant : 0.003819/°C

Substitute into equation 1

R = 8.6(1+0.003819[42-29])

R = 9.027 Ω.

Hence the resistance = 9.027 Ω

9.02Ω

The resistivity of a conductor increases as temperature increases. In this case, silver, which is a great conductor, will increase its resistivity linearly over a range of increasing temperatures . The relationship between resistance(R) and temperature(T) is given as;

R = R₀ (1 + α(T - T₀))     (i)

where;

R and R₀ are the final and initial resistances of the material (silver in this case)

α = temperature coefficient of resistivity of the material (silver) = 0.0038/°C

T and T₀ are the final and initial temperatures.

From the question;

R₀ = 8.60Ω

T₀ = 29.0°C

T = 42.0°C

Substitute these values into equation (i);

R = 8.60 (1 + 0.0038(42.0 - 29.0))

R = 8.60(1 + 0.0038(13))

R = 8.60(1 + 0.0494)

R = 8.60(1.0494)

R = 9.02Ω

Therefore, the resistance at 42.0°C is 9.02Ω

specific latent heat of fusion, lf, of a substance is defined as the amount of heat required to change a unit mass of the substance from solid to liquid state, without any change in the temperature.

explanation:

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