Question

On a winter day the temperature of the tap water is 20°C whereas the room temperature is 5°C. Water is stored in a tank of capacity 0.5 m3 for household use. If it were possible to use the heat liberated by the water to lift a 10 kg mass vertically, how high can it be lifted as the water comes to the room temperature? Take g = 10 m s−2.

Answer 1

Explanation:

Step 1:

Given:

$\mathrm{T}_{\mathrm{i}}=20^{\circ} \mathrm{C}$

Final temperature of the water (room temperature), $T_{t}=5^{\circ} \mathrm{C}$

$\begin{array}{l}{\Delta T=20^{\circ} \mathrm{C}-5^{\circ} \mathrm{C}=15^{\circ} \mathrm{C}} \\{\mathrm{V}=0.5 \mathrm{m}^{3}} \\{\mathrm{d}=1000 \mathrm{kg} / \mathrm{m}^{3}} \\{M=(0.5 \times 1000) \mathrm{kg}=500 \mathrm{kg}}\end{array}$

Where  

Ti is Initial temperature of the water

$\Delta T$ is Change in temperature

$V$is Volume of water

$\mathrm{d}$ is Density of water,

$M$ is Mass of the water,  

Step 2:

Energy released as the water temperature decreases from 20 ° C to 5 ° C

$\begin{aligned}&Q=M \times S \times \Delta T\\&Q=(500 \times 4200 \times 15) J\\&Q=(500 \times 4200 \times 15) J\\&Q=(75 \times 420 \times 1000) J\\&Q=31500 \times 1000 \mathrm{J}=315 \times 10^{5} \mathrm{J}\end{aligned}$

Step 3:

Let be h the height the mass is lifted to.

The energy needed to heighten the block = mgh = $10 \times 10 \times h=100 h$

$Q=m g h$

$100 h=315 \times 10^{5} J$

$\mathrm{h}=315 \times 10^{3} \mathrm{m}=315 \mathrm{km}$