JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 6)
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 ?
2.5 $$\times$$ 102 s
4.1 $$\times$$ 101 s
2.4 $$\times$$ 103 s
2.5 $$\times$$ 101 s
Explanation
$$\Delta$$Q = $$\Delta$$U + $$\Delta$$W
$${{\Delta Q} \over {\Delta t}} = {{\Delta U} \over {\Delta t}} + {{\Delta W} \over {\Delta t}}$$
$${{6000} \over {60}}{J \over {\sec }} = {{2.5 \times {{10}^3}} \over {\Delta t}} + 90$$
$$\Delta$$t = 250 sec
$${{\Delta Q} \over {\Delta t}} = {{\Delta U} \over {\Delta t}} + {{\Delta W} \over {\Delta t}}$$
$${{6000} \over {60}}{J \over {\sec }} = {{2.5 \times {{10}^3}} \over {\Delta t}} + 90$$
$$\Delta$$t = 250 sec
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