Topic: Thermodynamics
One hundredth of a kilogram of air is compressed In a piston-cylinder device. At an instant of time when $T=400K.$ the rate at which work is being done on the air is 8.165 kW, and heat is being removed at a rate of 1.0 kW, the rate of temperature rise will be -
- A. $10~K/s$
- B. $100~K/s$
- C. $1000~K/s$
- D. $10000~K/s$
Answer: C. $1000~K/s$
Explanation: Given, $m_{air}=\frac{1}{100}kg$, $\delta\dot{W}=-8.165~kW$ (work done on air), $\delta\dot{Q}=-1~kW$ (heat removed). According to first law of thermodynamics: $\delta\dot{Q}=d\dot{U}+\delta\dot{W} \Rightarrow -1=\frac{mC_{v}dT}{dt}-8.165 \Rightarrow \frac{mC_{v}dT}{dt}=7.165$. For air $C_{v}=0.718~kJ/kg~K$. So, $\frac{dT}{dt}=\frac{7.165}{0.01 \times 0.718}=1000~K/s$.