Question

Determine the work done by 2. 00 kg of water when it is all boiled to steam at 100 ∘c. Assume a constant pressure of 1. 00 atm.

Answer 1

The work done by the water at the constant pressure is $-4.52 \times 10^6 \ J$.

Heat required to boil the water

The heat required to boil the water is calculated as follows;

$Q = mL_v$

where;

• $L_v$ is the heat of vaporization of water = 22.6 x 10⁵ J/kg
• m is mass of water

$Q = 2 \times 22.6 \times 10^5\\\\Q = 4.52 \times 10^6 \ J$

Work done by the water

The work done by the water at the constant pressure is equal to the heat gained by the water.

$W = - \Delta H\\\\W = -4.52 \times 10^6 \ J$.

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