Question #136628
A faulty thermometer reads maelting point -10degree Celsius .it reads 60 degree c in place of 50 degree c .what is the temperature of boiling point of water on this scale ?
1
Expert's answer
2020-10-05T10:53:43-0400

The lower fixed point on the wrong scale is -10 ºC. Let n be the number of divisions between upper and lower fixed points on this scale. If Θ is the reading on this scale, then,

C0100=Θ(10)n\frac{C – 0}{100} = \frac{Θ-(-10)}{n}

When C = 50 ºC, Θ = 60 ºC

500100=60(10)n=70n\frac{50 – 0}{100} = \frac{60-(-10)}{n} = \frac{70}{n}

n=70×10050=140n = 70 \times \frac{100}{50} = 140

C0100=Θ(10)140\frac{C – 0}{100} = \frac{Θ-(-10)}{140}

On the Celsius scale, boiling point of water is 100 ºC

1000100=Θ(10)140\frac{100 – 0}{100} = \frac{Θ-(-10)}{140}

Θ = 140 – 10 = 130 ºC

Answer: 130 ºC

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