A platinum-resistance thermometer has a resistance of 5Ω at 0°C and 9Ω at l00°C. Assuming that resistance changes uniformly with temperature, calculate the ...

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A platinum-resistance thermometer has a resistance of 5 Ω at 0°C and 9 Ω at 100°C, and it is assumed that resistance changes uniformly with temperature. We need to calculate the resistance of the thermometer when the temperature is 45°C. We can start by finding the change in resistance as the temperature changes from 0°C to 100°C: ΔR = 9 Ω - 5 Ω = 4 Ω This change occurs over a temperature range of 100°C - 0°C = 100°C, so the resistance changes by 4 Ω per 100°C. To find the resistance at 45°C, we can use the following proportion: (ΔR / ΔT) = (R₂ - R₁) / (T₂ - T₁) Where: ΔR = change in resistance ΔT = change in temperature R₁ = initial resistance (at 0°C) R₂ = final resistance (at 45°C) T₁ = initial temperature (0°C) T₂ = final temperature (45°C) Substituting the known values, we get: (4 Ω / 100°C) = (R₂ - 5 Ω) / (45°C - 0°C) Simplifying the equation: R₂ - 5 Ω = (4 Ω / 100°C) x 45°C R₂ - 5 Ω = 1.8 Ω R₂ = 6.8 Ω Therefore, the resistance of the platinum-resistance thermometer at 45°C is 6.8 Ω. Answer: 6.8 Ω.