Question

An electric kettle raises the temperature of the certain mass of water from 0${}^{\circ }C$ to 100${}^{\circ }C$ in 2.1 minutes. It takes 11.2 minutes to boil off the water. Calculate the specific heat of vaporization.

• A. 5260 Jg^-1
• B. 3240 Jg^-1
• C. 2240 Jg^-1
• D. 2000 Jg^-1

Answer 1

The correct option is C

2240 Jg^-1

Let, mass of water be m and specific latent heat of vaporization = L

Heat absorbed by water in one second = Heat absorbed by steam in one second

$\frac{m\times C_w\times (100–0)}{Time}$ = $\frac{mL}{Time}$

$\frac{4.2\times 100}{2.1\times 60}$ = $\frac{L}{11.2\times 60}$

2 $\times$ 100 = $\frac{L}{11.2}$

L = 2240 Jg${}^{-1}$​