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?

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:

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:

   ...(1)

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

    and

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

So, the ratio of electrical power of heaters is:

    ....(2)

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

Substitute the suitable values in the above equation.

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator

the inner core.

good luck: )

klicker