Question

On a winter day the temperature of the tap water is 200C whereas the room temperature is 50C. Water is stored in a tank of capacity 0.5m3 for household use. If it were possible to use the heat liberated by the water to lift a 10kg mass vertically, how high can it be lifted as the water comes to the room temperature? Take g=10ms-

Answer 1

Hint
In this question, first we need to find out the heat liberated by the water for this we have formula i.e. $Q = M \times S \times \Delta T$, here as volume of water is given and density of water is $1000kg{m^{ - 3}}$, so we can find the value of mass after that we can find the value of Q. now, as the energy needed is in the form of potential energy so $Q = mgh$, from this we can find the value of height h.

Complete step by step answer
Here, it is given that, The initial temperature of the water is ${T_i} = {20^0}C$
Final temperature of the water (at room temperature) is ${T_f} = {5^0}C$
Change in temperature is $\Delta T = {20^0}C - {5^0}C = {15^0}C$
It is also given that
Volume of water is $V = 0.5{m^3}$
Density of water is $D = 1000kg{m^{ - 3}}$
As we know that $density = \dfrac{{mass}}{{volume}}$
$ \Rightarrow mass = density \times volume$
$ \Rightarrow M = 0.5 \times 1000 = 500kg$
Now, the heat liberated when the temperature of water changes from 200C to 50C is
$ \Rightarrow Q = M \times S \times \Delta T$
Where, Q is the heat liberated
S is the specific heat of the water i.e. $S = 4200Jk{g^{ - 1}}{K^{ - 1}}$
∆T is the change in temperature of water
On substituting the values in the formula, we get
$ \Rightarrow Q = \left( {500 \times 4200 \times 15} \right)J$
$ \Rightarrow Q = 31500 \times 1000J = 315 \times {10^5}J$ ……………………………….. (1)
Now, let the height to which the mass is lifted be h, then
The energy required to lift the block is = mgh
As the, mass of block lifted is 10 kg and acceleration due to gravity is g=10ms-2
The energy required to lift the block is $E = mgh = 10 \times 10 \times h$ ……………………… (2)
As, in the question it is given that the heat energy liberated is used in lift up the block, therefore, Using the equation (1) and (2), we get
$ \Rightarrow 100h = 315 \times {10^5}J$
$ \Rightarrow h = 315 \times {10^3}m$
$ \Rightarrow h = 315km$
Hence, this is the required height.

Note
Heat liberation means the amount of heat liberated when any chemical compound is formed. Specific heat is measured by the amount of heat energy required to raise one gram of one degree Celsius of the product. Water specific heat is 4.2 joule per gram per degree Celsius or 1 calorie per gram per degree Celsius.