5 X 10−3 − 3 kg of liquid at its boiling point is evaporated in 20s by the heat generated by a resistor of 2Ω Ω when a current of 10A is used. The specific ...

To solve this problem, we need to calculate the specific latent heat of vaporization of the liquid. The specific latent heat of vaporization, denoted as \(L\), is defined as the amount of heat required to convert 1 kilogram of a liquid into a gas at constant temperature and pressure. The formula for specific latent heat of vaporization is given by:

L = \(\frac{Q}{m}\)

Where:

• Q is the heat supplied to the liquid (in joules)
• m is the mass of the liquid that is evaporated (in kilograms)

First, we need to calculate the total heat energy \(Q\) generated by the resistor. The heat produced by an electrical resistor can be calculated using the formula:

Q = I^2Rt

Where:

• I is the current (in amperes)
• R is the resistance (in ohms)
• t is the time for which current is flowing (in seconds)

Given:

• I = 10 A
• R = 2 Ω
• t = 20 s

Substituting these values into the formula for Q:

Q = (10^2) * 2 * 20 = 100 * 2 * 20 = 4000 J

Now that we have the total heat energy supplied, let's calculate the specific latent heat of vaporization:

Given that the mass \(m\) of the liquid evaporated is \(5 \times 10^{-3}\) kg, we can substitute the values into the formula for \(L\):

L = \(\frac{4000}{5 \times 10^{-3}} = \frac{4000}{0.005} = 800,000 J/kg\)

Therefore, the specific latent heat of vaporization of the liquid is 8.0 x 10⁵ J/kg.