Question

1 kg of water is heated from 40°C to 70°C, if its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J kg–1 K–1)

• A. 2.44 × 105 J
• B. 1.62 × 105 J
• C. 1.24 × 105 J
• D. 2.62 × 105 J

Answer 1

Answer: (C)

1.24 × 105 J

Answer 2

: Since volume of water remains constant, then work done

$\Delta W = PdV = 0$

According to first pair of thermodynamics

}{= 1 \times 4148 \times (70 - 40) = 4148 \times 30 }{= 124440J = 1.244 \times 10^{5}J}$$