Respuesta :
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:
[tex]\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }[/tex]
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:
[tex]\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }[/tex] ...(1)
The relation to determine electrical power for both heaters A and B are:
[tex]P_{1}=\frac{V^{2} }{R_{1} }[/tex] and
[tex]P_{2}=\frac{V^{2} }{R_{2} }[/tex]
Here V is the voltage applied to both the heaters and is equal.
So, the ratio of electrical power of heaters is:
[tex]\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }[/tex] ....(2)
But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,
[tex]\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }[/tex]
Substitute the suitable values in the above equation.
[tex]\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }[/tex]
[tex]\frac{R_{1} }{R_{2} }=5.80[/tex]
