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

Question

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) 84.5 J (B) 104.3 J (C) 42.2 J (D) 208.7 J

Tardigrade Admin · Accepted Answer

Δ Q = Δ U + Δ W ⇒ 54 ×4.18 = Δ U + 1.013 ×105 (167.1 ×10-6 - 0) ⇒ Δ U = 208.7 J

Q. A sample of $0.1 g$ of water at $10 0^{\circ} C$ and normal pressure $(1.013 \times 1 0^{5} N m^{- 2})$ requires $54 c a l$ of heat energy to convert to steam at $10 0^{\circ} C$ . If the volume of the steam produced is $167.1 cc$ , the change in internal energy of the sample, is

84.5 J

104.3 J

42.2 J

208.7 J

$Δ Q = Δ U + Δ W$
$\Rightarrow 54 \times 4.18 = Δ U + 1.013 \times 1 0^{5} (167.1 \times 1 0^{- 6} - 0)$
$\Rightarrow Δ U = 208.7 J$

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