5 X 10 ^ -3 kg of liquid at its boiling point is evaporated in 20s by the heat generated by a resistor of 2…

Analysis

To solve this problem, we need to equate the electrical energy supplied by the resistor to the heat energy required to vaporize the liquid.

Concept

1. Electrical Energy ($H$): The heat generated by an electric current flowing through a resistor is given by Joule's Law: $H = I^2Rt$ Where: • $I$ = Current (in Amperes) • $R$ = Resistance (in Ohms) • $t$ = Time (in seconds)

2. Heat of Vaporization ($Q$): The heat energy required to change a substance from liquid to gas at its boiling point is: $Q = mL_v$ Where: • $m$ = Mass (in kg) • $L_v$ = Specific latent heat of vaporization (in Jkg$^{-1}$)

Since the electrical energy is used to evaporate the liquid, we set $H = Q$.

Step-by-Step Solution

Step 1: Calculate the electrical energy ($H$) Given:

• $I = 10\text{ A}$
• $R = 2\text{ }\Omega$
• $t = 20\text{ s}$

$H = I^2Rt$ $H = (10)^2 \times 2 \times 20$ $H = 100 \times 2 \times 20$ $H = 4000\text{ J}$

Step 2: Calculate the specific latent heat ($L_v$) Given:

• $m = 5 \times 10^{-3}\text{ kg}$
• $H = 4000\text{ J}$

Using $H = mL_v$: $4000 = (5 \times 10^{-3}) \times L_v$

$L_v = \frac{4000}{5 \times 10^{-3}}$ $L_v = \frac{4000}{0.005}$ $L_v = 800,000\text{ Jkg}^{-1}$

Step 3: Convert to standard form $800,000 = 8.0 \times 10^5\text{ Jkg}^{-1}$

Conclusion

Comparing our result with the given options, we find that the calculated value matches option B.

Exam Tip: Always check your units before calculating! In JAMB Physics, mass is almost always expected in kilograms (kg). If the question had given the mass in grams, you would have needed to divide by 1000 first to avoid a common pitfall.