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

Respuesta :

Answer:

[tex]\Delta U=3997.63J[/tex]

Explanation:

According to the first law of thermodynamics:

[tex]\Delta U=Q-W[/tex]

Here [tex]\Delta U[/tex] 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:

[tex]W=P\Delta V[/tex]

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

[tex]W=P(0)\\W=0[/tex]

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

[tex]\Delta U=Q[/tex]

1 cal is equal to 4.186 J, so:

[tex]\Delta U=955cal*\frac{4.186J}{1cal}\\\Delta U=3997.63J[/tex]