Question

An experiment takes $10$ min to raise the temperature of water from $0$ to $100$ and another $55$ min to convert it totally into steam by a stabilized heater. The latent heat of vaporization comes out to be
A. $530\,cal\,{g^{ - 1}}$
B. $540\,cal\,{g^{ - 1}}$
C. $550\,cal\,{g^{ - 1}}$
D. $560\,cal\,{g^{ - 1}}$

Answer 1

For stabilized heaters, the rate of heat supply will be constant. Calculate the amount of heat transferred by the heater in $10$ minutes. This much amount of heat will be required to raise the temperature of water. Calculate the amount of heat supplied by the heater in $55$ minutes. This much heat will be required to convert water into steam. Using both equations, find the required quantity.

Complete step by step answer:
Assuming the heater is supplying heat at a constant rate of $Q\,cal\,{s^{ - 1}}$ . The total heat supplied by the heater in $10$ minutes will be
$10 \times 60 \times Q\,cal\,{s^{ - 1}}$
Converting minutes to seconds.
This much heat will be required to raise the temperature of water from $0$ to $100$ . The amount of energy required to raise $\Delta T$ temperature of water having mass $m$ is given as
$mc\Delta T$
Here, $c$ is the specific heat capacity of water.
The amount of heat supplied by the heater is equal to the amount of energy required to raise the temperature of water by $\Delta T$ . Thus, both the quantities must be equal:
$mc\Delta T = 600Q$
$\Rightarrow mc(100 - 0) = 600Q$
$\Rightarrow mc \times 100 = 600Q$
$\Rightarrow mc = 6Q$ --equation $1$
The amount of heat supplied by heater in $55$ minutes will be
$55 \times 60 \times Q\,cal\,{s^{ - 1}}$
Converting minutes to seconds.
The amount of energy required to convert water having mass $m$ to steam is given as:
$mL$
Here $L$ is the latent heat of vaporization.
The amount of heat supplied by the heater in $55$ minutes must be equal to the amount of energy required to convert water having mass $m$ to steam. Thus, both the quantities must be equal:
$\Rightarrow mL = 3300Q$
From equation $1$ , we have $m = \dfrac{{6Q}}{c}$
Substituting this value in above equation, we get
$\dfrac{{6Q}}{c}L = 3300Q$
$\Rightarrow L = \dfrac{{3300}}{6} \times c$
For water $c = 1\,cal\,{g^{ - 1\,}}{C^0}^{\, - 1}$
$\Rightarrow L = \dfrac{{3300}}{6} \times (1)$
$\Rightarrow L = 550\,cal\,{g^{ - 1}}$
Thus, the latent heat of vaporization comes out to be $550\,cal\,{g^{ - 1}}$ .

So, the correct answer is “Option C”.

Note:
The specific heat capacity of water is $c = 1\,cal\,{g^{ - 1\,}}{C^0}^{\, - 1}$ .
As, there was no heat loss in the system therefore, the entire heat given by the heater was used either to raise the temperature of the water or to convert water into steam.
Convert minutes into seconds as the SI unit of time is seconds.