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The contents of a well-insulated tank are heated by a resistor of 23 Ω in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy (∆U) during the process (kW) are
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- Q = 0, W = –2.3, ∆U = +2.3
- Q = +2.3, W = 0, ∆U = +2.3
- Q = –2.3, W = 0, ∆U = –2.3
- Q = 0, W = + 2.3, ∆U = –2.3
Correct Option: A
Welectric = i2R 
Welectric = i2R
= (102 × 20) watts = – 2.3 kW (on system)
First law: Q – w = ∆U
∴ 0 – (– Welectric) = ∆u
