An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90 W

Q: Why do we divide the First Law equation by $\Delta t$?

The standard First Law equation ($\Delta Q = \Delta U + \Delta W$) deals with absolute amounts of energy. However, the problem provides 'power' (which is work done per unit time) and a heat supply rate (Joules per minute). Dividing the equation by $\Delta t$ converts it into a 'rate equation', allowing us to plug in the given values directly.

Q: Why did we convert 6000 J/min to 100 J/s?

In physics, consistency in units is critical. Power is given in Watts, which means Joules per second (J/s). If we mix J/min with J/s in the same equation, our final answer will be completely wrong. Converting everything to standard SI units (J/s) prevents this.

Q: What does 'power delivered' mean in the context of the First Law?

'Power delivered' by the system means the system is doing work on its surroundings. In the First Law equation, the rate of work done ($\frac{\Delta W}{\Delta t}$) is exactly what power is. Since the system is doing the work, it is taken as a positive value on the right side of the equation.

Q: Can we solve this without converting to rates first?

Yes, you can! You could express total heat as $Q = (100 \text{ J/s}) \times t$ and total work as $W = (90 \text{ J/s}) \times t$. Plugging these into $Q = \Delta U + W$ gives $100t = 2500 + 90t$. Solving for $t$ yields the exact same result. The rate equation is just a slightly faster way to write it.

Q: What if the system was doing work ON the gas instead of delivering power?

If work was being done ON the system (e.g., compressing a gas), the work done BY the system ($\Delta W$) would be negative. The equation would then look like $\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} - P$. This means both the heat added and the work done on the system would contribute to increasing its internal energy.

Question

An electric appliance supplies $6000	ext{ J/min}$ heat to the system. If the system delivers a power of $90	ext{ W}$. How long it would take to increase the internal energy by $2.5 	imes 10^3	ext{ J}$?

Options:

(A) $2.5 	imes 10^2	ext{ s}$

(B) $4.1 	imes 10^1	ext{ s}$

(C) $2.4 	imes 10^3	ext{ s}$

(D) $2.5 	imes 10^1	ext{ s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a thermodynamic system—perhaps a specialized engine or a sealed gas chamber. Energy is constantly flowing in and out of this system.

We are told that an electric appliance is pumping heat into the system at a rate of $6000 	ext{ J/min}$. Simultaneously, the system isn't just sitting idle; it is actively doing work on its surroundings, delivering a power of $90 	ext{ W}$. Our goal is to find out exactly how long it will take for the system's internal energy to build up and increase by $2.5 	imes 10^3 	ext{ J}$.

### The Master Equation

To connect heat, work, and internal energy, we rely on the bedrock of thermodynamics: **The First Law of Thermodynamics**.

Mathematically, it is expressed as:
$$ \Delta Q = \Delta U + \Delta W $$

However, the data provided in the problem isn't in total amounts of energy, but rather in *rates* (energy per unit time). To adapt our master equation, we simply divide every term by the time interval $\Delta t$:
$$ \frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t} $$

This modified equation tells us that the rate at which heat enters the system equals the rate at which internal energy increases, plus the rate at which the system does work (which is the definition of power).

### Unit Conversion: The Silent Trap

Before we rush to substitute numbers, we must ensure all units are consistent. The standard SI unit for power and energy rates is the Watt ($	ext{W}$), which is equivalent to Joules per second ($	ext{J/s}$).

The power delivered is already in standard units:
$$ P = \frac{\Delta W}{\Delta t} = 90 	ext{ W} = 90 	ext{ J/s} $$

But the heat supply rate is given in Joules per minute. We must convert this:
$$ \frac{\Delta Q}{\Delta t} = 6000 	ext{ J/min} = \frac{6000}{60} 	ext{ J/s} = 100 	ext{ J/s} $$

### Final Calculation

Now, we are ready to substitute our pristine, unit-matched values into the rate equation. We know the target change in internal energy is $\Delta U = 2.5 	imes 10^3 	ext{ J}$.

$$ 100 = \frac{2.5 	imes 10^3}{\Delta t} + 90 $$

Subtracting $90$ from both sides gives us the net rate at which energy is being stored inside the system:
$$ 10 = \frac{2.5 	imes 10^3}{\Delta t} $$

Finally, isolating $\Delta t$:
$$ \Delta t = \frac{2.5 	imes 10^3}{10} $$
$$ \Delta t = 2.5 	imes 10^2 	ext{ s} $$

It will take exactly $250$ seconds for the internal energy to reach the desired level. This perfectly matches option (a).

Animated Solution for Physics - Thermodynamics: An electric appliance supplies $6000 J/min$ heat to the system. If the system delivers a power of $90 W$ . How long it would take to increase the internal energy by $2.5 \times 1 0^{3} J$ ?

Select Answer:

Visualized Solution

\text{System Setup}

\text{First Law of Thermodynamics}

\text{Given Rates and Unit Conversion}

\text{Substituting Values}

\text{Solving for } \Delta t

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Unit Conversion

Final Calculation

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