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A faulty thermometer reads $$5°$$C in melting ice and $$95°$$C in steam. The correct temperature on absolute scale will be ______ K when the faulty thermometer reads $$41°$$C.
Correct Answer: 313
Temperature scales are constructed linearly based on two fixed reference points: the lower fixed point (melting point of ice) and the upper fixed point (boiling point of water/steam).
For any linear thermometer scale, the ratio of the difference between a reading and the ice point to the total interval between the steam point and the ice point remains perfectly constant:
$$\frac{\text{Reading} - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{Constant}$$
Where:
Let us compare the faulty thermometer scale ($$W$$) with the standard Celsius scale ($$C$$):
Equating their linear scale ratios:
$$\frac{W - \text{LFP}_W}{\text{UFP}_W - \text{LFP}_W} = \frac{C - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}$$
$$\frac{W - 5}{95 - 5} = \frac{C - 0}{100 - 0}$$
$$\frac{W - 5}{90} = \frac{C}{100}$$
We are given that the faulty thermometer reads $$W = 41^\circ\text{C}$$. Substituting this value into our equation:
$$\frac{41 - 5}{90} = \frac{C}{100}$$
$$\frac{36}{90} = \frac{C}{100}$$
Simplify the fraction on the left side:
$$\frac{2}{5} = \frac{C}{100}$$
$$C = \frac{2}{5} \times 100 = 40^\circ\text{C}$$
The relationship between the standard Celsius scale and the absolute thermodynamic Kelvin scale ($$T$$) is defined by:
$$T = C + 273$$
Substituting the true calculated value of $$C = 40^\circ\text{C}$$:
$$T = 40 + 273 = 313 \,\, \text{K}$$
Correct Numerical Answer: 313
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