Question

An electric immersion heater of 1.08kW is immersed in water. After the water has reached a temperature of ${{100}^{0}}C$, how much time will be required to produce 100g of steam?
A. 50s
B. 420s
C. 105s
D. 210s

Answer 1

Hint: The problem is based on the thermodynamics concept of latent heat vaporization, which is equal to 540 calories and the basic conversion from energy to work done, using which time taken can be found out. The heat energy is given by an immersion rod.

Complete step-by-step answer:
Let’s start by understanding what Latent heat of vaporization is. Latent heat of vaporization of water, is the amount of heat energy required by 1g of water to be converted to 1g of steam. For water, Latent of vaporization = 540cal =\[\left( 540\times 4.2 \right)J\]= 2268J. Hence, 2268 Joules of energy is required to heat 1g of water to 1g of steam. This implies, to convert 100gm of water to steam, the amount of energy required will be\[100\times 2268\]= 226800J.
Now, we know the amount of energy that needs to be given to 100g of water is 226800J. That amount of energy is given by the immersion rod. The power of the immersion rod is given as 1.08kW. That means,$1.08\times {{10}^{3}}J$, of work is done by the immersion rod per second. The work done is used in heating the water.
Energy is equal to work done per unit time, hence, the amount of time required by the immersion rod to convert 100g of water to 100g of steam is,$226800=t(1.08\times {{10}^{3}})\Rightarrow t=\dfrac{226800}{1.08\times {{10}^{3}}}\Rightarrow t=\dfrac{22680}{108}\Rightarrow t\approx 210s.$
Hence, the immersion rod can convert 100g of water into 100g of steam in 210s.
Therefore the correct option is C.

Note: It’s important to know how to convert energy into work and work done into energy. Energy=(Work done)/time taken to do the work. Now, to know that the conversion is happening correctly, you should also know the units of energy and work. Units of work are, wattage short for watt, denoted as W. 1W=1J/s. Units of energy are well known as joules (or) calories. In the problem, I refer to the work done by the immersion heater as power, as that is how you will read it for any electrical appliance. It’s written over the electrical appliances as power rating of 1.08kW.