Question

Apply the first law of thermodynamics to a resistor carrying a current i. Identify which of the quantities ∆Q, ∆U and ∆W are zero, positive and negative.

Answer 1

Answer:

work done is +ve and heat energy is -ve.ande change in internal energy is zero.

Explanation:

Beacause the heat liberates when the current pass through it so change in heat energy is -ve

Answer 2

Positive work is performed by the battery on a resistor with the current i. Thus, ∆W is positive. To increase thermal energy, the work performed on the resistor is used. Thus, ∆Q is positive. There will be an increase in the temperature of the resistor, so ∆U is positive.

Explanation:

According to the first law of thermodynamics, the isolated system’s total energy is constant.  

where ,

ΔU is the internal energy’s change,

∆Q is the system heat, and  

∆W is the work performed.

• The battery is now performing positively on a resistor with the current i. So, ∆W is positive.
• On a resistor, the work performed is used to raise the thermal energy as heat, thus ∆Q is positive. If the resistor’s temperature increases, then ΔU is positive.