Question

Two grams of water are sealed in a rigid container; then the water is vaporized by heating. 955 cal of heat is needed for vaporization. What is the change in internal energy of the water? Answer in units of J

Answer 1

`\Delta U=3997.63J`

Step-by-step explanation:

According to the first law of thermodynamics:

`\Delta U=Q-W`

Here
`\Delta U` is the change in the internal energy of the system, Q is the supplied heat to the system and W is the work done by the system. The work done by the system is defined as:

`W=P\Delta V`

Where P is the presion and
`\Delta V` is the volume change. The volume remains constant, since the water are sealed in a rigid container:

`W=P(0)\\W=0`

So, the change in internal energy of the water is:

`\Delta U=Q`

1 cal is equal to 4.186 J, so:

`\Delta U=955cal*(4.186J)/(1cal)\\\Delta U=3997.63J`