A sample of 0.1 g of water at 100°C and normal pressure $(1.013 × 10^{5}Nm^{-2})$ 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
42.2 J
208.7 J
104.3 J
84.5 J
Explanation:
$\triangle Q=\triangle U+\triangle W$
$\Rightarrow 54 \times 4.18 =\triangle U +1.013 \times 10^{5}(167.1 \times 10^{-6}-0)$
$\Rightarrow \triangle U=208.7 J$
$\triangle Q=\triangle U+\triangle W$
$\Rightarrow 54 \times 4.18 =\triangle U +1.013 \times 10^{5}(167.1 \times 10^{-6}-0)$
$\Rightarrow \triangle U=208.7 J$
