1. Tardigrade
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  3. Physics
  4. 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

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Q. A sample of $0.1\,g$ of water at $100^{\circ}C $ and normal pressure $(1.013 \times 10^5 \, Nm^{ -2})$ requires $54\,cal$ of heat energy to convert to steam at $100^{\circ}C $. If the volume of the steam produced is $167.1\, cc$, the change in internal energy of the sample, is

NEETNEET 2018Thermodynamics

Solution:

$\Delta Q = \Delta U + \Delta W $
$\Rightarrow 54 \times4.18 = \Delta U + 1.013 \times10^{5} \left(167.1 \times10^{-6} - 0\right) $
$\Rightarrow \Delta U = 208.7\, J $