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JEE MAIN 2023
08-04-2023 S1
Question
When a 60 W electric heater is immersed in a gas for 100s in a constant volume container with adiabatic walls, the temperature of the gas rises by 5°C. The heat capacity of the given gas is ___J K–1 (Nearest integer)
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Solution
Sol. $$ \begin{aligned} & \text { Power of heater }=60 \mathrm{~W} \\ & =60 \mathrm{~J} / \mathrm{sec} \\ & \text { Total energy emitted }=60 \times 100=6000 \mathrm{~J} \\ & \text { Heat capacity } \times \text { temp difference }=6000 \\ & \text { Heat capacity }=\frac{6000}{5}=1200 \mathrm{JK}^{-1} \end{aligned} $$
Question Tags
JEE Main
Chemistry
Easy
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