Q1. 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 OP = -2.2067x10-12 (0ª – 81×10*) dt 0(0) = 1200K where 0 is in K and t in seconds. Find the temperature at t=480 seconds using the Runge-Kutta 4th order method. Assume a step size of h=120 seconds.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Q1. 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
de
:-2.2067x10-12 (e* -81×10* )
dt
0(0) =1200K
where 0 is in K and t in seconds. Find the temperature at t=480 seconds using the Runge-Kutta 4th order method.
Assume a step size of h=120 seconds.
II
Transcribed Image Text:Q1. 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 de :-2.2067x10-12 (e* -81×10* ) dt 0(0) =1200K where 0 is in K and t in seconds. Find the temperature at t=480 seconds using the Runge-Kutta 4th order method. Assume a step size of h=120 seconds. II
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