On finding your stove out of order, you decide to boil the water for a cup of tea by shaking it in a thermos flask. Suppose that you use tap water at , the water falls $\mathbf{32}\mathbf{}\mathbf{\text{cm}}$each shake, and you make 27 shakes each minute. Neglecting any loss of thermal energy by the flask, how long (in minutes) must you shake the flask until the water reaches$\mathbf{100}\mathbf{\text{°C}}$ ?

1. The initial temperature of tap water is $T_i=19\text{°Cor}292\text{K}$
2. The water falls from a height with each shake of$h=32\text{cmor}0.32\text{m}$
3. The number of shakes each minute is $N/t=27/\text{min}$
4. The final temperature of water is $T_f=100\text{°Cor}373\text{K}$.

The amount of heat required to increase the temperature of 1 gram of a substance by 1 degree Celsius is known as its specific heat. We can use the concept of specific heat of water. The energy required to raise the temperature of water up to its boiling point can be supplied by the gravitational potential energy during the shaking of the flask.

Formulae:

Heat transferred by the body,$Q=cm(T_f-T_i)$…(i)

where c is specific heat capacity of water = $4190\frac{\text{J}}{\text{kgK}}$

Potential energy of the body,$U=mgh$. …(ii)

Let m be the mass of water andbe the specific heat of water.

During each shake, it reaches at a heighthence, it gets gravitational potential energy. This energy can be used for raising the temperature of the water. Hence, using equation (ii), the heat is given by

$Q=mgh$

There are a number of shakes, then

$Q=Nmgh$

Letbe the time taken to raise the temperature of water up to boiling point. Thus, the value of heat energy radiated per time is given by

$\frac{Q}{t}=\frac{Nmgh}{t}$ …(iii)

The number of shakes per minute is$\frac{N}{t}=27/\text{min}$.

Solving equation (iii) further, we can get the time required to get boiling water as

$\begin{array}{c}t=\frac{cm(T_f-T_i)}{(N/t)mgh}\text{(fromequation(i))}\\ =\frac{c(T_f-T_i)}{(N/t)gh}\\ =\frac{4190\frac{\text{J}}{\text{kgK}}(373\text{K}-292\text{K})}{\frac{\text{27}}{\text{min}}\times 9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\times 0.32\text{m}}\text{(substitutingthegivenvalues)}\\ =4.01\times {10}^3\text{min}\\ \approx 4.0\times {10}^3\text{min}\end{array}$

Hence, the value of the required time is $4.0\times {10}^3\text{min}$.