A loaded domestic refrigerator consumes 1.0 kW-hr of energy during a certain period. The internal energy of the system drops by 5000 kJ. The quantity of heat transferred from the system to the surrounding will be
8.6 MJ
Step 1 – Electrical work input: \(W_{\mathrm{in}}=1\ \mathrm{kWh}=3600\ \mathrm{kJ}\).
Step 2 – Energy balance: With heat rejected taken as positive, \(\Delta U=W_{\mathrm{in}}-Q_{\mathrm{out}}\). Hence \(Q_{\mathrm{out}}=3600-(-5000)=8600\ \mathrm{kJ}=8.6\ \mathrm{MJ}\).
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:
