Question

The mass of a closed system is 100 kg with an initial velocity of 5 m/s. During some process, Its velocity increases to 15 m/s, its elevation rises by 20 m, The system receives 40 kJ of heat and 5 kJ of work. If the system produces 0.0025 kWhr of electrical energy. What is the change in internal energy of the system ?

• A. 4.8 kJ
• B. 5.6 kJ
• C. 6.4 kJ
• D. 7.0 kJ

Answer 1

The correct option is C 6.4 kJ

$\Delta PE=mg(h_2-h_1)=100\times 9.81\times \left(20\right)=19620J$
$\Delta KE=\frac{1}{2}(v_2^2-v_1^2)=\frac{1}{2}\times 100({15}^2-5^2)=10000J$
$Q_{1-2}=40000J$

$W_{1-2}=W_{\text{out}}=W_{\text{in}}=0.0025\times 1000\times 3600-5000=4000J$

From Ist law of thermodynamics

$Q_{1-2}=W_{1-2}+\Delta E$

$Q_{1-2}=W_{1-2}+\Delta U+\delta KE+\Delta PE$

$\Delta U=40000-4000-19620-10000$

$6380J=6.38kJ\simeq 6.4kJ$