Back to Directory
NEET Medium

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

A

42.2 J

B

42.2 J

C

104.3 J

D

84.5 J

Step-by-Step Solution

ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W

⇒54×4.18=ΔU+1.013×105(167.1×10−6−0)\Rightarrow 54 \times 4.18 = \Delta U + 1.013 \times 10^5(167.1 \times 10^{-6} - 0)

⇒ΔU=208.7 J\Rightarrow \Delta U = 208.7 \text{ J}

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started