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?

Options

• 4.1 × 10¹s

• 2.4 × 10³s

• 2.5 × 10²s

• 2.5 × 10¹s

Answer 1

2.5 × 10²s

Explanation:

Given `(DeltaQ)/(Deltat) = 6000J/min = 6000/60J/s`

`(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

`(ΔU)/(Δt) = (ΔQ)/(Δt) - (ΔW)/(Δt)`

Now putting the supplied values,

`(2.5 xx 10^3)/(Deltat) = 100 - 90 = 10`

Δt = `(2.5 xx 10^3)/10`

Δt = 2.5 × 10²s