An electric kettle can heat certain quantity of water from$19°\text{C}$ to$100°\text{C}$ in 6 minutes. Time it takes to boil the same quantity of water into steam at $100°\text{C}$ $\left({\text{L}}_{\text{steam}}=540\text{cal/g}\right)$  is

$\Delta \text{t}=81°\text{C}$                 Let the mass of water = m gram

$\text{S}={\text{1calg}}^{-1}{\text{C}}^{-1}$

$\text{Q}=\text{ms}\Delta \text{t}=\text{m}\left(1\right)\times 81$

$\therefore \text{Q}=81\text{m}\Rightarrow 6\min$

Heat required to convert same quantity of water at $100°\text{C}$ into steam at$100°\text{C}$

$\text{Q}=540\text{m}\left(\text{L}=540\text{cal/g}\right)$  

We get

$81\text{m}\text{cal}-6\min$

$540\text{m}-?$

Time required$=\frac{\text{540}\text{m}}{\text{81}\text{m}}\times 6=40\min$