A ball at 1200 K is allowed to cool down in air at an ambient temperature of 300 K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by --2.2067×10-124 -81×108 ). y0)= 1200K, x=0 (240) 240 dx where y is in K and x in seconds. Find the temperature at t=240 using Runge-Kutta 4th order method. Assuming a step size h=240 seconds.

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A ball at 1200 K is allowed to cool down in air at an ambient temperature of 300 K.
Assuming heat is lost only due to radiation, the differential equation for the temperature
of the ball is given by
dy
:-2.2067x10-12(y4 -81×108). y(0)=1200K, x= 0 (240) 240
dx
where y is in K and x in seconds. Find the temperature at t=240 using Runge-Kutta 4th
order method. Assuming a step size h=240 seconds.
Transcribed Image Text:A ball at 1200 K is allowed to cool down in air at an ambient temperature of 300 K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by dy :-2.2067x10-12(y4 -81×108). y(0)=1200K, x= 0 (240) 240 dx where y is in K and x in seconds. Find the temperature at t=240 using Runge-Kutta 4th order method. Assuming a step size h=240 seconds.
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