Question

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 × 10³ J?

पर्याय

• 4.1 × 10¹s

• 2.4 × 10³s

• 2.5 × 10²s

• 2.5 × 10¹s

Answer 1

2.5 × 10²s

Explanation:

Given ${\displaystyle \frac{\Delta Q}{\Delta t}=6000\frac{J}{min}=\frac{6000}{60}\frac{J}{s}}$

${\displaystyle \frac{dW}{dt}}$ = 90 W,

Let Δt internal energy increase by ΔU over time.

So, ΔU = 2.5 × 10³ J

According to the first law of thermodynamics,

ΔU = Q - W

${\displaystyle \frac{\Delta U}{\Delta t}=\frac{\Delta Q}{\Delta t}-\frac{\Delta W}{\Delta t}}$

Now putting the supplied values,

${\displaystyle \frac{2.5\times {10}^3}{\Delta t}=100-90=10}$

Δt = ${\displaystyle \frac{2.5\times {10}^3}{10}}$

Δt = 2.5 × 10²s