1 kg liquid (specific heat = 3.0 kJ/kg-K) is stirred in closed chamber and its temperature is raised by 10°C. Heat loss to the surrounding is 3.0 kJ. The work done on the water during the process will be:
33 kJ
The liquid gains \(\Delta U=mc\Delta T=1(3)(10)=30\ \mathrm{kJ}\). Stirring must supply this plus the 3 kJ heat loss, so \(W_{on}=30+3=33\ \mathrm{kJ}\).
Important Questions from First Law of Thermodynamics
Heat transfer in a cyclic process are +20 kJ, -5 kJ, -10 kJ and +15kJ. Net work done for this cycle will be given by:
An open system is one in which ______.
A system will be in thermodynamic equilibrium only if it is in ____.
A. Thermal equilibrium
B. Mechanical equilibrium
C. Chemical equilibrium
The gas constant (R) is equal to the _____ of two specific heats.
In a reversible adiabatic process the ration \(\left( {\frac{{{T_1}}}{{{T_2}}}} \right)\) is equal to:
