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 ___________ J K⁻¹ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Adiabatic walls and fixed volume send all the heater's energy into the gas: divide it by the rise.

1. Heat supplied. Power is energy per second, so the heater delivers q = P × t = 60 W × 100 s = 6000 J.
2. Where it goes. The walls are adiabatic, so no heat leaks out. The volume is constant, so the gas does no expansion work. All 6000 J goes into the gas and raises its temperature.
3. Temperature change. A rise of 5 °C is the same as a rise of 5 K, because both scales have the same size of degree.

   C = q/(Δ T) = (6000 J)/(5 K)

4. So. C = 6000 / 5 = 1200 J K⁻¹, so the answer is 1200.

The common slip: 22. Turning the 5 °C rise into 278 K gives 6000/278 ≈ 22, not 1200. Add 273 only to an actual temperature, never to a temperature change.

The idea to keep. Heat capacity is heat absorbed per kelvin of temperature rise. For a heater, heat is power times time, and a change in temperature has the same value in kelvin as in degrees Celsius.

Answer from the NTA final answer key. Solution by PrepWiser.