Question

Two immersion heaters, A and B, are both connected to a 120.0-V supply. Heater A can raise the temperature of 1.00 L of water from 20.0°C to 90.0°C in 5.00 min, whereas heater B can raise the temperature of 5.80 L of water from 20.0°C to 90.0°C in 5.00 min. What is the ratio of the resistance of heater A to the resistance of heater B?

Answer 1

Answer:

Ratio of resistance of heater A to resistance of heater B is 5.80

Explanation:

Consider C be the specific heat of water, R₁ and R₂ be the resistance of heater A and heater B respectively.

Given:

Mass of water in heater A, m₁ = 1 L

Mass of water in heater B, m₂ = 5.80 L

Initial temperature, T₀ = 20 ⁰C

Final temperature, T₁ = 90 ⁰C

Time, t = 5 min

Amount of heat required to raise the temperature of water by heaters A and B are given by:

Q₁ = m₁C(T₁ - T₀)       and  

Q₂ = m₂C(T₁ - T₀)

Ratio of power used by both the heaters A and B is:

$\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }$

Since, time t, temperature difference(T₁ - T₀) and specific heat C are same for both the heaters A and B. So, the above equation becomes:

$\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }$    ...(1)

The relation to determine electrical power for both heaters A and B are:

$P_{1}=\frac{V^{2} }{R_{1} }$     and

$P_{2}=\frac{V^{2} }{R_{2} }$

Here V is the voltage applied to both the heaters and is equal.

So, the ratio of electrical power of heaters is:

$\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }$     ....(2)

But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,

$\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }$

Substitute the suitable values in the above equation.

$\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }$

$\frac{R_{1} }{R_{2} }=5.80$