A faulty thermometer reads 5^{circ}C in melti | vedclass

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(C) Let $T_f$ be the reading on the faulty thermometer and $T_c$ be the reading on the Celsius scale.The formula for conversion is $\frac{T_f - LFP}{U

Vedclass · Accepted Answer

$313$

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(C) Let $T_f$ be the reading on the faulty thermometer and $T_c$ be the reading on the Celsius scale.The formula for conversion is $\frac{T_f - LFP}{UFP - LFP} = \frac{T_c - 0}{100 - 0}$,where $LFP$ is the Lower Fixed Point and $UFP$ is the Upper Fixed Point.Given $LFP = 5^{\circ}C$ and $UFP = 95^{\circ}C$,we have $\frac{41 - 5}{95 - 5} = \frac{T_c}{100}$.$\frac{36}{90} = \frac{T_c}{100}$.$T_c = \frac{36}{90} 	imes 100 = 40^{\circ}C$.To convert to the absolute scale (Kelvin),$T_K = T_c + 273.15 \approx 40 + 273 = 313 \, K$.

$A$ faulty thermometer reads $5^{\circ}C$ in melting ice and $95^{\circ}C$ in steam. The correct temperature on the absolute scale will be $.........K$ when the faulty thermometer reads $41^{\circ}C$.

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