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 ?

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 ? (A) 2.5 × 102 s (B) 4,1 × 101 s (C) 2.4 × 103 s (D) 2.5 × 101 s

Tardigrade Admin · Accepted Answer

Δ Q =Δ U +Δ W (Δ Q /Δ t )=(Δ U /Δ t )+(Δ W /Δ t ) (6000/60) ( J / sec )=(2.5 × 103/Δ t )+90 Δ t =250 sec

Q. 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 $2.5 \times 1 0^{3} J$ ?

$2.5 \times 1 0^{2} s$

$4, 1 \times 1 0^{1} s$

$2.4 \times 1 0^{3} s$

$2.5 \times 1 0^{1} s$

$Δ Q = Δ U + Δ W$
$\frac{Δ Q}{Δ t} = \frac{Δ U}{Δ t} + \frac{Δ W}{Δ t}$
$\frac{6000}{60} \frac{J}{s e c} = \frac{2.5 \times 1 0 ^{3}}{Δ t} + 90$
$Δ t = 250 sec$

Q. An electric appliance supplies 6000J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5×103J ?

2.5×102s

4,1×101s

2.4×103s

2.5×101s