Question

A stone of 20 kg mass and a tank containing 200 kg water comprise a

system. The stone is 15 m above the water level initially. The stone and water are at the same

temperature initially. If the stone falls into water, then determine ∆U, ∆PE, ∆KE, Q and W, when

(i) the stone is about to enter the water,

(ii) the stone has come to rest in the tank, and

(iii) the heat is transferred to the surroundings in such an amount that the stone and water

come to their initial temperature.

Answer 1

Solution:

Given;

Mass of stone;$m_s=20kg$

Mass of water in tank; $m_w=200kg$

Height of stone above water level; =15m

Applying the first law of thermodynamics;

$Q=(U_2-U_1)+m(\frac{C_2^2-C_1^2}{2})+mg(Z_2-Z_1)+W$

$Q=\Delta U+\Delta KE+\Delta PE+W...(1)$

Here ;

Q =Heat leaving the boundary.

Hence;

(i)the stone is about to enter the water;

Q=0

W=0

$\Delta U$ =0

And;

$-\Delta KE=\Delta PE=mg(Z_2-Z_1)$ =20×9.81×(0-15)=-2943J

$\Delta KE=2943J$

$\Delta PE=-2943J$

(ii)Stone comes to rest in the tank;

Q=0

W=0

$\Delta KE=0$

$\Delta PE=-2943J$

By substitution into (1),we have;

0=$\Delta U+0+\Delta PE+0$

$\Delta U=-\Delta PE=2943J$

$\Delta U=2943J$

(iii)When stone and water come to their initial temperature;

W=0

$\Delta KE =0$

Using equation (1),we get;

Q=$-\Delta U=-2943J$

Negative signs shows that heat is lost from the system to the surrounding.