Question

An electric appliance supplies $6000\text{J/min}$ of heat to the system. If the system delivers power of $90\text{W}$, how long would it take to increase its internal energy by $2.5\times {10}^3\text{J}$?

• A. $2.5\times {10}^1\text{s}$
  
• B. $4.1\times {10}^1\text{s}$
  
• C. $2.4\times {10}^3\text{s}$
  
• D. $2.5\times {10}^2\text{s}$
  

Answer 1

The correct option is D $2.5\times {10}^2\text{s}$


Here, $Q=6000\text{J/min}=100\text{J/s}$

$P=\frac{W}{t}=90\text{W}=90\text{J/s}$
$\Delta U=2.5\times {10}^3\text{J}$

Let the internal energy increases in a time interval $t$

According to first law of thermodynamics,

$Q=\Delta U+W$

$\because Q,W$ are in $\text{Watt}$
$100=\frac{\Delta U}{t}+90$
$\therefore t=\frac{\Delta U}{10}=\frac{2.5\times {10}^3}{10}=2.5\times {10}^2\text{s}$

Hence, $\left(B\right)$ is the correct answer.