Thermometry and Calorimetry | JEE Main previous year questions with Solutions

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Topic Question Set

Q 11 :

On a temperature scale ‘X’, the boiling point of water is 65°X and the freezing point is –15°X. Assume that the X scale is linear. The equivalent temperature corresponding to –95°X on the Fahrenheit scale would be [2023]

–63°F
–112°F
–48°F
–148°F

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3

4

(4)

$\frac{X - X_{freez}}{X_{boil} - X_{freez}} = \frac{t - 32}{212 - 32}$

$\frac{- 95 - (- 15)}{65 - (- 15)} = \frac{t - 32}{180}$

$\Rightarrow \frac{- 80}{80} = \frac{t - 32}{180}$

$t = - 180 + 32$

$\Rightarrow t = - 148 ° F$

Q 12 :

A faulty thermometer reads 5°C in melting ice and 95°C in steam. The correct temperature on the absolute scale will be _____ K when the faulty thermometer reads 41°C. [2023]

0%

(313)

$\frac{41 ° - 5 °}{95 ° - 5 °} = \frac{C - 0 °}{100 ° - 0 °}$

$\Rightarrow C = \frac{36}{90} \times 100 = 40 ° C = 313 K$

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