Question

Determine the change in internal energy of 1.00 kg of Water when it's all boiled to steam at 100⁰C. Assume a constant pressure of 1.00 atm. (1.00 kg of steam at 100⁰C has a volume of 1.67 m³)

a) 2.26 × 10^6 J
b) 3.53 × 10^5 J
c) 2.26 × 10^5 J
d) 3.53 × 10^6 J

Answer 1

The change in internal energy of 1.00 kg of water when it boils to steam at 100°C is 2.26 × 10^6 J.

Step-by-step explanation:

The change in internal energy of 1.00 kg of water when it boils to steam at 100°C can be calculated using the formula:

Change in Internal Energy = Heat added - Work done

Since the question states that the volume of 1.00 kg of steam at 100°C is 1.67 m³, we can calculate the work done using the formula:

Work done = Pressure * Change in Volume

Given that the pressure is 1.00 atm and the change in volume is 1.67 m³, the work done is 1.00 atm * 1.67 m³ = 1.67 J. Therefore, the change in internal energy is 2.26 × 10^6 J - 1.67 J = 2.26 × 10^6 J. Option a) is the correct answer.