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 $2.5 \times 10^3$ J ?

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

Hint:

Ensure all quantities are in standard SI units (Watts or Joules/second) before performing calculations. 1 J/min is 1/60 W.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The first law of thermodynamics states that the heat added to a system (\(\Delta Q\)) is equal to the sum of the change in internal energy (\(\Delta U\)) and the work done by the system (\(\Delta W\)).
Step 2: Key Formula or Approach:
\[ \frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt} \]
where \(\frac{dQ}{dt}\) is the rate of heat supply and \(\frac{dW}{dt}\) is the power output.
Step 3: Detailed Explanation:
1. Rate of heat supply:
\[ \frac{dQ}{dt} = 6000 \text{ J/min} = \frac{6000}{60} \text{ J/s} = 100 \text{ W} \]
2. Power delivered by the system:
\[ \frac{dW}{dt} = 90 \text{ W} \]
3. Rate of increase of internal energy:
\[ \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} = 100 - 90 = 10 \text{ J/s} \]
4. Time taken to increase internal energy by \(\Delta U = 2.5 \times 10^3 \text{ J}\):
\[ t = \frac{\Delta U}{dU/dt} = \frac{2.5 \times 10^3 \text{ J}}{10 \text{ J/s}} = 250 \text{ s} = 2.5 \times 10^2 \text{ s} \]
Step 4: Final Answer:
The time required is $2.5 \times 10^2$ s.