Verified Solution • 2025–26 Session moderate

QJEEM21Phbffa: JEE Main 2021 Physics - Class 11 JEEM21Phbffa Solution: An electric appliance supplies 6000 J/min heat to the system.…

An electric appliance supplies 6000 J/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 ×\times 103 J ?

Multiple Choice Options

  • A 2.5 ×\times 102 s
  • B 4.1 ×\times 101 s
  • C 2.4 ×\times 103 s
  • D 2.5 ×\times 101 s

Step-by-Step Verified Solution

Δ\Delta Q = Δ\Delta U + Δ\Delta W

ΔQΔt=ΔUΔt+ΔWΔt{{\Delta Q} \over {\Delta t}} = {{\Delta U} \over {\Delta t}} + {{\Delta W} \over {\Delta t}}

600060Jsec⁡=2.5×103Δt+90{{6000} \over {60}}{J \over {\sec }} = {{2.5 \times {{10}^3}} \over {\Delta t}} + 90

Δ\Delta t = 250 sec

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