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 ?
Solution
$\Delta$Q = $\Delta$U + $\Delta$W<br><br>$${{\Delta Q} \over {\Delta t}} = {{\Delta U} \over {\Delta t}} + {{\Delta W} \over {\Delta t}}$$<br><br>$${{6000} \over {60}}{J \over {\sec }} = {{2.5 \times {{10}^3}} \over {\Delta t}} + 90$$<br><br>$\Delta$t = 250 sec
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.