HELP PLEASE! ASAP

Question

A Nichrome heater dissipates 501.6 W when the applied potential difference is 110 V and

the wire temperature is 800°C. Find the dissipation rate (power) assuming the wire temperature

was held at 200°C by immersing the wire in a bath of cooling oil. The applied potential

difference remains the same, and temperature coefficient for Nichrome is 4x10^-4 (°C)^-1 at 800°C.

PLEASE SOLVE WITH STEPS.

Answer 1

Power is given as P =V²/R

P = Power

V = Potential difference

R = Resistance

P₁= Power at 800⁰C

P2= Power at 200⁰C

Also R_t = R₀ ( 1+α t)

Rt =Resistance at t⁰C and R₀ is Resistance at 0⁰C

P₁/P₂= V1²/V2² (1+ α*200)/( 1+α*800)= (1+ α*200)/( 1+α*800) ,as V₁ = V₂ = 110 Volts

501.6/P₂ = (1+ α*200)/( 1+α*800)

Now find P2 by Plugging in α=4*10^-4