Q23 / 90 · JEE Main 2023 (08 Apr Shift 1) · Chemistry

When a $60 W$ electric heater is immersed in a gas for $100 s$ 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 ${JK}^{- 1}$ (Nearest integer)

Correct answer: 1200

Solution

$Heat capacity = \frac{Heat absorbed}{change in temperature}$
1 Watt = 1 Joule per second
Hence, heat absorbed $= 60 \times 100 J$
Change in temperature $= 5 ° C$
Heat capacity $= \frac{60 \times 100 J}{5}$
$= 1200 {JK}^{- 1}$

Hence, the correct answer is 1200.

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When a 60 W electric heater is immersed in a gas for 100 s 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 JK-1 (Nearest integer)

Solution

Relevant Questions

When a $60 W$ electric heater is immersed in a gas for $100 s$ 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 ${JK}^{- 1}$ (Nearest integer)