Question

A cup of water at 5°C (system) is placed in a microwave oven and the oven is turned on for one minute during which the water begins to boil. Which of the following option is true ?

• A. q = +ve, w = 0, ΔU = -ve
• B. q = +ve, w = -ve, ΔU = +ve
• C. q = +ve, w = -ve, ΔU = -ve
• D. q = -ve, w = -ve, ΔU = -ve

Hint:

In thermodynamics, always remember the IUPAC convention: energy into the system is positive ($q>0$), and work done by the system is negative ($w<0$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We apply the First Law of Thermodynamics: \(\Delta U = q + w\). We must identify the signs of heat (\(q\)), work (\(w\)), and internal energy (\(\Delta U\)) based on the process described.
Step 2: Detailed Explanation:
1. Heat (\(q\)): The microwave provides energy to the water to increase its temperature. Energy enters the system, so \(q = +ve\).
2. Internal Energy (\(\Delta U\)): The temperature of the water rises from 5°C to 100°C (boiling). Since temperature is a measure of internal energy for water, \(\Delta U = +ve\).
3. Work (\(w\)): As the water heats up and begins to boil, it expands against the atmospheric pressure (vapor formation). Work is done by the system on the surroundings, so \(w = -ve\).
Step 3: Final Answer:
The correct option is q = +ve, w = -ve, $\Delta$ U = +ve.