Question

An experiment takes $10$ min to raise temperature of water from $0$ and $100$ and another $55$ min to convert it totally into steam by a stabilized heater. The latent heat of vaporization comes out to be

• A. $530cal/g$
• B. $540cal/g$
• C. $550cal/g$
• D. $560cal/g$

Answer 1

Answer: (C)

$550cal/g$

Assumption: Specific heat capacity of ware is $s=1cal{}^o{C}^{-1}{g}^{-1}$

Let the input power be $P$.

Heat supplied in $10min$ is:

$H=Pt=600P............\left(i\right)$

Given, this raises the temperature by ${100}^{o}C$

$H=ms\Delta T=100m...........\left(ii\right)$ where $m$ is in grams

From (i) and (ii), we get:

$600P=100m..............\left(iii\right)$

Now, heat supplied in the next $55min$ is

$H_2=55\times 60P=3300P............\left(iv\right)$

From the definition of latent heat

$H_2=mL............\left(v\right)$

From (iv) and (v),

$3300P=mL...........\left(vi\right)$

From (v) and (vi),

$L=550cal/g$